Chapter 8 Introduction To Trigonometry And Its Applications

Solution:

We need to find the value of $(\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ)$.

We know the values of the standard trigonometric angles:

$\sin 30^\circ = \frac{1}{2}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\cos 60^\circ = \frac{1}{2}$

Substitute these values into the given expression:

$(\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ) = \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)$

Now, simplify the expression:

$\left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) = \frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - \frac{1}{2}$

Combine the terms:

$\left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) = 0 + 0 = 0$

Thus, the value of $(\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ)$ is $0$.

The correct option is (B).

Solution:

We need to find the value of $\frac{\tan 30^\circ}{\cot 60^\circ}$.

We know the values of the standard trigonometric angles:

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

$\cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}}$

Substitute these values into the given expression:

$\frac{\tan 30^\circ}{\cot 60^\circ} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}}$

Simplify the expression:

$\frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 1$

Thus, the value of $\frac{\tan 30^\circ}{\cot 60^\circ}$ is $1$.

The correct option is (D).

Solution:

We need to find the value of $(\sin 45^\circ + \cos 45^\circ)$.

We know the values of the standard trigonometric angles:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

Substitute these values into the given expression:

$(\sin 45^\circ + \cos 45^\circ) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$

Combine the terms:

$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{1+1}{\sqrt{2}} = \frac{2}{\sqrt{2}}$

Simplify the expression by rationalizing the denominator or by recognizing that $2 = \sqrt{2} \times \sqrt{2}$:

$\frac{2}{\sqrt{2}} = \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}} = \sqrt{2}$

Thus, the value of $(\sin 45^\circ + \cos 45^\circ)$ is $\sqrt{2}$.

The correct option is (B).

Solution:

Given:

$\cos A = \frac{4}{5}$

To Find:

The value of $\tan A$.

Consider a right-angled triangle ABC, right-angled at B, with respect to angle A.

We know that $\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$.

Given $\cos A = \frac{4}{5}$, let the adjacent side (AB) be $4k$ and the hypotenuse (AC) be $5k$, where $k$ is a positive number.

Using the Pythagoras theorem in $\triangle$ABC:

$(\text{Hypotenuse})^2 = (\text{Adjacent side})^2 + (\text{Opposite side})^2$

$AC^2 = AB^2 + BC^2$

$(5k)^2 = (4k)^2 + BC^2$

$25k^2 = 16k^2 + BC^2$

$BC^2 = 25k^2 - 16k^2$

$BC^2 = 9k^2$

$BC = \sqrt{9k^2} = 3k$ (Since BC is a length, we take the positive square root)

Now we can find the value of $\tan A$. We know that $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}}$.

$\tan A = \frac{BC}{AB} = \frac{3k}{4k}$

$\tan A = \frac{3}{4}$

Alternate Solution (Using trigonometric identity):

We know the identity $\sin^2 A + \cos^2 A = 1$.

Substitute the given value of $\cos A$:

$\sin^2 A + \left(\frac{4}{5}\right)^2 = 1$

$\sin^2 A + \frac{16}{25} = 1$

$\sin^2 A = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}$

$\sin A = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$

Assuming angle A is acute (as typically in these problems unless otherwise specified), $\sin A$ is positive.

So, $\sin A = \frac{3}{5}$.

Now, use the identity $\tan A = \frac{\sin A}{\cos A}$.

$\tan A = \frac{\frac{3}{5}}{\frac{4}{5}}$

$\tan A = \frac{3}{5} \times \frac{5}{4}$

$\tan A = \frac{3}{4}$

The value of $\tan A$ is $\frac{3}{4}$.

The correct option is (B).

Solution:

Given:

$\sin A = \frac{1}{2}$

To Find:

The value of $\cot A$.

Consider a right-angled triangle ABC, right-angled at B, with respect to angle A.

We know that $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}}$.

Given $\sin A = \frac{1}{2}$, let the opposite side (BC) be $1k$ and the hypotenuse (AC) be $2k$, where $k$ is a positive number.

Using the Pythagoras theorem in $\triangle$ABC:

$(\text{Hypotenuse})^2 = (\text{Adjacent side})^2 + (\text{Opposite side})^2$

$AC^2 = AB^2 + BC^2$

$(2k)^2 = AB^2 + (1k)^2$

$4k^2 = AB^2 + k^2$

$AB^2 = 4k^2 - k^2$

$AB^2 = 3k^2$

$AB = \sqrt{3k^2} = \sqrt{3}k$ (Since AB is a length, we take the positive square root)

Now we can find the value of $\cot A$. We know that $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}}$.

$\cot A = \frac{AB}{BC} = \frac{\sqrt{3}k}{1k}$

$\cot A = \sqrt{3}$

Alternate Solution (Using trigonometric identity):

We know the identity $\sin^2 A + \cos^2 A = 1$.

Substitute the given value of $\sin A$:

$\left(\frac{1}{2}\right)^2 + \cos^2 A = 1$

$\frac{1}{4} + \cos^2 A = 1$

$\cos^2 A = 1 - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4}$

$\cos A = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$

Assuming angle A is acute (as typically in these problems unless otherwise specified), $\cos A$ is positive.

So, $\cos A = \frac{\sqrt{3}}{2}$.

Now, use the identity $\cot A = \frac{\cos A}{\sin A}$.

$\cot A = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$

$\cot A = \frac{\sqrt{3}}{2} \times \frac{2}{1}$

$\cot A = \sqrt{3}$

The value of $\cot A$ is $\sqrt{3}$.

The correct option is (A).

Solution:

We are asked to find the value of the expression:

$[\text{cosec} (75^\circ + \theta) – \sec (15^\circ – \theta) – \tan (55^\circ + \theta) + \cot (35^\circ – \theta)]$

We use the complementary angle identities:

$\sec (90^\circ - x) = \text{cosec} x$

$\cot (90^\circ - x) = \tan x$

Consider the first two terms: $\text{cosec} (75^\circ + \theta) – \sec (15^\circ – \theta)$.

Notice that $(75^\circ + \theta) + (15^\circ – \theta) = 75^\circ + 15^\circ + \theta - \theta = 90^\circ$.

So, $(15^\circ – \theta)$ and $(75^\circ + \theta)$ are complementary angles.

Using the identity $\sec (90^\circ - x) = \text{cosec} x$, we can write:

$\sec (15^\circ – \theta) = \sec (90^\circ - (75^\circ + \theta)) = \text{cosec} (75^\circ + \theta)$

Therefore, the first part of the expression becomes:

$\text{cosec} (75^\circ + \theta) – \sec (15^\circ – \theta) = \text{cosec} (75^\circ + \theta) – \text{cosec} (75^\circ + \theta) = 0$

Now consider the last two terms: $-\tan (55^\circ + \theta) + \cot (35^\circ – \theta)$.

Notice that $(55^\circ + \theta) + (35^\circ – \theta) = 55^\circ + 35^\circ + \theta - \theta = 90^\circ$.

So, $(35^\circ – \theta)$ and $(55^\circ + \theta)$ are complementary angles.

Using the identity $\cot (90^\circ - x) = \tan x$, we can write:

$\cot (35^\circ – \theta) = \cot (90^\circ - (55^\circ + \theta)) = \tan (55^\circ + \theta)$

Therefore, the second part of the expression becomes:

$-\tan (55^\circ + \theta) + \cot (35^\circ – \theta) = -\tan (55^\circ + \theta) + \tan (55^\circ + \theta) = 0$

The value of the entire expression is the sum of these two parts:

$[\text{cosec} (75^\circ + \theta) – \sec (15^\circ – \theta) – \tan (55^\circ + \theta) + \cot (35^\circ – \theta)] = 0 + 0 = 0$

The value of the expression is $\mathbf{0}$.

The correct option is (B).

Solution:

Given:

$\sin \theta = \frac{a}{b}$

To Find:

The value of $\cos \theta$.

We use the fundamental trigonometric identity:

$\sin^2 \theta + \cos^2 \theta = 1$

Substitute the given value of $\sin \theta$ into the identity:

$\left(\frac{a}{b}\right)^2 + \cos^2 \theta = 1$

$\frac{a^2}{b^2} + \cos^2 \theta = 1$

Solve for $\cos^2 \theta$:

$\cos^2 \theta = 1 - \frac{a^2}{b^2}$

$\cos^2 \theta = \frac{b^2 - a^2}{b^2}$

Take the square root of both sides to find $\cos \theta$:

$\cos \theta = \pm \sqrt{\frac{b^2 - a^2}{b^2}}$

$\cos \theta = \pm \frac{\sqrt{b^2 - a^2}}{\sqrt{b^2}}$

Assuming $\theta$ is an acute angle or in a quadrant where $\cos \theta$ is positive, and $b > 0$, we take the positive root:

$\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}$

Alternatively, using a right-angled triangle, if $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{b}$, let the opposite side be $a$ and the hypotenuse be $b$ (assuming $a>0, b>0$).

By Pythagoras theorem, the adjacent side = $\sqrt{b^2 - a^2}$.

Then $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{b^2 - a^2}}{b}$.

Comparing this result with the given options, we find the correct option.

The correct option is (C).

Solution:

Given:

$\cos (\alpha + \beta) = 0$

To Find:

A simplified expression for $\sin (\alpha – \beta)$.

We are given that $\cos (\alpha + \beta) = 0$. This means that $(\alpha + \beta)$ is an angle whose cosine is 0.

The general solution for $\cos x = 0$ is $x = (2n+1)\frac{\pi}{2}$, where $n$ is an integer.

So, $\alpha + \beta = (2n+1)\frac{\pi}{2}$.

For simplicity, and as suggested by the options provided, let us consider the principal case where $n=0$, which gives $\alpha + \beta = \frac{\pi}{2}$ (or $90^\circ$).

From $\alpha + \beta = 90^\circ$, we can express $\alpha$ in terms of $\beta$:

$\alpha = 90^\circ - \beta$

Now substitute this expression for $\alpha$ into the expression we want to simplify, $\sin (\alpha – β)$:

$\sin (\alpha – β) = \sin ((90^\circ - \beta) – \beta)$

Simplify the argument of the sine function:

$(90^\circ - \beta) – \beta = 90^\circ - 2\beta$

So, $\sin (\alpha – β) = \sin (90^\circ - 2\beta)$.

Using the complementary angle identity $\sin (90^\circ - x) = \cos x$, we can write:

$\sin (90^\circ - 2\beta) = \cos (2\beta)$

Therefore, if $\cos (\alpha + \beta) = 0$ (considering the principal case $\alpha + \beta = 90^\circ$), $\sin (\alpha – \beta)$ can be reduced to $\cos (2\beta)$.

The correct option is (B) $\cos 2β$.

Solution:

We need to find the value of the product $\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 89^\circ$.

We can group the terms in pairs using the complementary angle identity: $\tan (90^\circ - x) = \cot x$.

Also, we know that $\cot x = \frac{1}{\tan x}$.

Consider the terms in the product. The angles range from $1^\circ$ to $89^\circ$. We can pair angles that sum up to $90^\circ$:

$(1^\circ, 89^\circ), (2^\circ, 88^\circ), \ldots, (44^\circ, 46^\circ)$.

There is one term left in the middle: $\tan 45^\circ$.

Let's look at a pair, for example, $\tan 1^\circ$ and $\tan 89^\circ$:

$\tan 89^\circ = \tan (90^\circ - 1^\circ) = \cot 1^\circ$

So, $\tan 1^\circ \times \tan 89^\circ = \tan 1^\circ \times \cot 1^\circ = \tan 1^\circ \times \frac{1}{\tan 1^\circ} = 1$.

This applies to every pair of angles $(x, 90^\circ - x)$ for $x$ from $1^\circ$ to $44^\circ$:

$\tan x \times \tan (90^\circ - x) = \tan x \times \cot x = \tan x \times \frac{1}{\tan x} = 1$.

The product can be written as:

$(\tan 1^\circ \tan 89^\circ) (\tan 2^\circ \tan 88^\circ) \cdots (\tan 44^\circ \tan 46^\circ) \tan 45^\circ$

Substituting the value 1 for each pair product and the value of $\tan 45^\circ$:

We know that $\tan 45^\circ = 1$.

So the product becomes:

$(1) \times (1) \times \cdots \times (1) \times 1$

There are 44 such pairs, and the term $\tan 45^\circ$.

The product is $1^{44} \times 1 = 1 \times 1 = 1$.

The value of (tan1° tan2° tan3° ... tan 89°) is $\mathbf{1}$.

The correct option is (B).

Solution:

Given:

$\cos 9\alpha = \sin \alpha$

$9\alpha < 90^\circ$

To Find:

The value of $\tan 5\alpha$.

We use the complementary angle identity $\sin x = \cos (90^\circ - x)$.

Using this identity, we can rewrite $\sin \alpha$ as $\cos (90^\circ - \alpha)$.

Substitute this into the given equation:

$\cos 9\alpha = \cos (90^\circ - \alpha)$

Since $9\alpha < 90^\circ$ and $\alpha$ must be positive (for $\sin \alpha$ to be involved in a standard trigonometric context as implied by the problem, and $\alpha = (90^\circ - 9\alpha)/10$, which is positive if $9\alpha < 90^\circ$), both $9\alpha$ and $90^\circ - \alpha$ are acute angles. If the cosine of two acute angles is equal, then the angles must be equal.

Therefore,

$9\alpha = 90^\circ - \alpha$

Now, solve for $\alpha$:

$9\alpha + \alpha = 90^\circ$

$10\alpha = 90^\circ$

$\alpha = \frac{90^\circ}{10}$

$\alpha = 9^\circ$

Now we need to find the value of $\tan 5\alpha$. Substitute the value of $\alpha$:

$5\alpha = 5 \times 9^\circ = 45^\circ$

So we need to find $\tan 45^\circ$.

We know that the value of $\tan 45^\circ$ is 1.

$\tan 5\alpha = \tan 45^\circ = 1$

The value of $\tan 5\alpha$ is $\mathbf{1}$.

The correct option is (C).

Solution:

Given:

$\triangle$ABC is right-angled at C.

To Find:

The value of $\cos (A + B)$.

In any triangle, the sum of the angles is $180^\circ$.

For $\triangle$ABC, the sum of angles A, B, and C is $180^\circ$.

$A + B + C = 180^\circ$

Since the triangle is right-angled at C, the angle C is $90^\circ$.

Substitute the value of C into the equation:

$A + B + 90^\circ = 180^\circ$

Solve for $A + B$:

$A + B = 180^\circ - 90^\circ$

$A + B = 90^\circ$

Now we need to find the value of $\cos (A + B)$. Substitute the value of $A + B$:

$\cos (A + B) = \cos 90^\circ$

We know that the value of $\cos 90^\circ$ is 0.

$\cos (A + B) = 0$

The value of $\cos (A + B)$ is $\mathbf{0}$.

The correct option is (A).

Solution:

Given:

$\sin A + \sin^2 A = 1$

To Find:

The value of the expression $(\cos^2 A + \cos^4 A)$.

From the given equation, we can rearrange the terms to get:

$\sin A = 1 - \sin^2 A$

We know the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$.

From this identity, we can write $\cos^2 A = 1 - \sin^2 A$.

Comparing the rearranged given equation with the identity, we see that:

$\sin A = \cos^2 A$

Now, consider the expression we need to evaluate: $(\cos^2 A + \cos^4 A)$.

We can rewrite $\cos^4 A$ as $(\cos^2 A)^2$.

So, the expression is $\cos^2 A + (\cos^2 A)^2$.

Substitute the relationship $\cos^2 A = \sin A$ into this expression:

$\cos^2 A + (\cos^2 A)^2 = (\sin A) + (\sin A)^2 = \sin A + \sin^2 A$

We are given that $\sin A + \sin^2 A = 1$.

Therefore, the value of the expression $(\cos^2 A + \cos^4 A)$ is equal to 1.

The value of $(\cos^2 A + \cos^4 A)$ is $\mathbf{1}$.

The correct option is (A).

Solution:

Given:

$\sin \alpha = \frac{1}{2}$

$\cos \beta = \frac{1}{2}$

To Find:

The value of $(\alpha + \beta)$.

We need to find the angles $\alpha$ and $\beta$ for which the given trigonometric conditions hold. Assuming $\alpha$ and $\beta$ are acute angles (in the range $0^\circ$ to $90^\circ$) as is common in basic trigonometry problems unless specified otherwise.

For $\sin \alpha = \frac{1}{2}$, the standard angle is $\alpha = 30^\circ$.

$\sin 30^\circ = \frac{1}{2}$

For $\cos \beta = \frac{1}{2}$, the standard angle is $\beta = 60^\circ$.

$\cos 60^\circ = \frac{1}{2}$

Now, we calculate the sum $(\alpha + \beta)$:

$\alpha + \beta = 30^\circ + 60^\circ$

$\alpha + \beta = 90^\circ$

The value of $(\alpha + \beta)$ is $\mathbf{90^\circ}$.

The correct option is (D).

Solution:

We need to evaluate the expression: $\left[ \frac{\sin^2 22^\circ \;+\; \sin^2 68^\circ}{\cos^2 22^\circ \;+\; \cos^2 68^\circ} + \sin^2 63^\circ + \cos 63^\circ\sin 27^\circ \right]$.

Let's evaluate the first part of the expression: $\frac{\sin^2 22^\circ \;+\; \sin^2 68^\circ}{\cos^2 22^\circ \;+\; \cos^2 68^\circ}$.

We use the complementary angle identity: $\sin (90^\circ - x) = \cos x$ and $\cos (90^\circ - x) = \sin x$.

Notice that $22^\circ + 68^\circ = 90^\circ$.

So, $\sin 68^\circ = \sin (90^\circ - 22^\circ) = \cos 22^\circ$.

And, $\cos 68^\circ = \cos (90^\circ - 22^\circ) = \sin 22^\circ$.

Substitute these into the first part:

Numerator: $\sin^2 22^\circ + \sin^2 68^\circ = \sin^2 22^\circ + (\cos 22^\circ)^2 = \sin^2 22^\circ + \cos^2 22^\circ$.

Using the identity $\sin^2 x + \cos^2 x = 1$, the numerator is $1$.

Denominator: $\cos^2 22^\circ + \cos^2 68^\circ = \cos^2 22^\circ + (\sin 22^\circ)^2 = \cos^2 22^\circ + \sin^2 22^\circ$.

Using the identity $\sin^2 x + \cos^2 x = 1$, the denominator is $1$.

So, the first part of the expression is $\frac{1}{1} = 1$.

Now, let's evaluate the second part of the expression: $\sin^2 63^\circ + \cos 63^\circ\sin 27^\circ$.

We use the complementary angle identity: $\sin (90^\circ - x) = \cos x$.

Notice that $63^\circ + 27^\circ = 90^\circ$.

So, $\sin 27^\circ = \sin (90^\circ - 63^\circ) = \cos 63^\circ$.

Substitute this into the second part:

$\sin^2 63^\circ + \cos 63^\circ\sin 27^\circ = \sin^2 63^\circ + \cos 63^\circ \times \cos 63^\circ$

$\sin^2 63^\circ + \cos 63^\circ\sin 27^\circ = \sin^2 63^\circ + \cos^2 63^\circ$.

Using the identity $\sin^2 x + \cos^2 x = 1$, this part is $1$.

Now, combine the values of the two parts to find the value of the entire expression:

Value = (First part) + (Second part)

Value = $1 + 1 = 2$

The value of the expression is $\mathbf{2}$.

The correct option is (B).

Solution:

Given:

$4 \tan \theta = 3$

To Find:

The value of $\left( \frac{4\sin\theta \;-\; \cos\theta}{4\sin\theta \;+\; \cos\theta} \right)$.

From the given equation, we can find the value of $\tan \theta$:

$4 \tan \theta = 3$

$\tan \theta = \frac{3}{4}$

We need to evaluate the expression $\frac{4\sin\theta \;-\; \cos\theta}{4\sin\theta \;+\; \cos\theta}$.

We can divide the numerator and the denominator of this expression by $\cos \theta$. This is valid as long as $\cos \theta \neq 0$. If $\cos \theta = 0$, then $\tan \theta$ would be undefined, which contradicts the given $4 \tan \theta = 3$. Therefore, $\cos \theta \neq 0$.

Dividing numerator and denominator by $\cos \theta$:

$\frac{4\sin\theta \;-\; \cos\theta}{4\sin\theta \;+\; \cos\theta} = \frac{\frac{4\sin\theta}{\cos\theta} \;-\; \frac{\cos\theta}{\cos\theta}}{\frac{4\sin\theta}{\cos\theta} \;+\; \frac{\cos\theta}{\cos\theta}}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we get:

$\frac{4\tan\theta \;-\; 1}{4\tan\theta \;+\; 1}$

Now, substitute the value of $\tan \theta = \frac{3}{4}$ into this expression:

$\frac{4\left(\frac{3}{4}\right) \;-\; 1}{4\left(\frac{3}{4}\right) \;+\; 1} = \frac{3 \;-\; 1}{3 \;+\; 1}$

Simplify the expression:

$\frac{3 \;-\; 1}{3 \;+\; 1} = \frac{2}{4} = \frac{1}{2}$

The value of the expression is $\mathbf{\frac{1}{2}}$.

The correct option is (C).

Solution:

Given:

$\sin \theta – \cos \theta = 0$

To Find:

The value of $(\sin^4 \theta + \cos^4 \theta)$.

From the given condition, we can write:

$\sin \theta = \cos \theta$

Since $\sin \theta = \cos \theta$, we can divide both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$. If $\cos \theta = 0$, then $\sin \theta$ would also be 0 from the given condition, which is not possible for any angle $\theta$).

$\frac{\sin \theta}{\cos \theta} = 1$

$\tan \theta = 1$

For acute angles, $\tan \theta = 1$ implies $\theta = 45^\circ$.

In this case, $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$.

Substituting these values into the expression $(\sin^4 \theta + \cos^4 \theta)$:

$(\sin 45^\circ)^4 + (\cos 45^\circ)^4 = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4$

Calculate the powers:

$\left(\frac{1}{\sqrt{2}}\right)^4 = \left(\left(\frac{1}{\sqrt{2}}\right)^2\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

So, $\sin^4 45^\circ = \frac{1}{4}$ and $\cos^4 45^\circ = \frac{1}{4}$.

Substitute these values back into the expression:

$\sin^4 \theta + \cos^4 \theta = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

Alternate Solution:

From $\sin \theta = \cos \theta$ and the identity $\sin^2 \theta + \cos^2 \theta = 1$, we can substitute $\cos \theta = \sin \theta$ into the identity:

$\sin^2 \theta + (\sin \theta)^2 = 1$

$2\sin^2 \theta = 1$

$\sin^2 \theta = \frac{1}{2}$

Since $\sin \theta = \cos \theta$, we also have $\cos^2 \theta = (\sin \theta)^2 = \sin^2 \theta = \frac{1}{2}$.

Now consider the expression $(\sin^4 \theta + \cos^4 \theta)$. We can write this as $(\sin^2 \theta)^2 + (\cos^2 \theta)^2$.

Substitute the value $\sin^2 \theta = \frac{1}{2}$ and $\cos^2 \theta = \frac{1}{2}$:

$(\sin^2 \theta)^2 + (\cos^2 \theta)^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

The value of $(\sin^4 \theta + \cos^4 \theta)$ is $\mathbf{\frac{1}{2}}$.

The correct option is (C).

Solution:

We need to find the value of $\sin (45^\circ + \theta) – \cos (45^\circ – \theta)$.

We use the complementary angle identity: $\cos x = \sin (90^\circ - x)$.

Consider the term $\cos (45^\circ – \theta)$. Let $x = 45^\circ – \theta$.

Then $90^\circ - x = 90^\circ - (45^\circ – \theta) = 90^\circ - 45^\circ + \theta = 45^\circ + \theta$.

Using the identity, $\cos (45^\circ – \theta) = \sin (90^\circ - (45^\circ – \theta)) = \sin (45^\circ + \theta)$.

Substitute this result back into the original expression:

$\sin (45^\circ + \theta) – \cos (45^\circ – \theta) = \sin (45^\circ + \theta) – \sin (45^\circ + \theta)$

Simplifying the expression gives:

$\sin (45^\circ + \theta) – \sin (45^\circ + \theta) = 0$

The value of the expression is $\mathbf{0}$.

The correct option is (B).

Solution:

Given:

Height of the pole = $6$ m

Length of the shadow = $2\sqrt{3}$ m

To Find:

The Sun's elevation angle.

Let the height of the pole be represented by the vertical side AB and the length of the shadow by the horizontal side BC of a right-angled triangle ABC, where $\angle$B is $90^\circ$. The Sun's elevation angle is the angle formed at the end of the shadow on the ground to the top of the pole, which is $\angle$BCA. Let $\theta$ be the angle of elevation.

In the right-angled triangle ABC, we have:

Opposite side to $\theta$ (Height of pole AB) = $6$ m

Adjacent side to $\theta$ (Length of shadow BC) = $2\sqrt{3}$ m

We can use the tangent trigonometric ratio, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle:

$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$

$\tan \theta = \frac{AB}{BC}$

Substitute the given values into the equation:

$\tan \theta = \frac{6}{2\sqrt{3}}$

Simplify the fraction:

$\tan \theta = \frac{\cancel{6}^3}{\cancel{2}_{1}\sqrt{3}}$

$\tan \theta = \frac{3}{\sqrt{3}}$

To rationalize the denominator, multiply the numerator and the denominator by $\sqrt{3}$:

$\tan \theta = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\tan \theta = \frac{3\sqrt{3}}{3}$

Simplify further:

$\tan \theta = \sqrt{3}$

We know from the standard trigonometric values that $\tan 60^\circ = \sqrt{3}$.

So, $\tan \theta = \tan 60^\circ$.

Therefore, $\theta = 60^\circ$.

The Sun's elevation is $\mathbf{60^\circ}$.

The correct option is (A).

False

Justification:

We need to check if the value of $\sin \theta + \cos \theta$ is always greater than 1 for all values of $\theta$.

Let's consider some values of $\theta$:

If $\theta = 0^\circ$, then $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$.

So, $\sin 0^\circ + \cos 0^\circ = 0 + 1 = 1$.

Since $1$ is not greater than $1$, the statement that $\sin \theta + \cos \theta$ is always greater than 1 is false.

Also consider $\theta = 90^\circ$:

$\sin 90^\circ = 1$ and $\cos 90^\circ = 0$.

So, $\sin 90^\circ + \cos 90^\circ = 1 + 0 = 1$.

Again, $1$ is not greater than $1$.

In fact, the maximum value of $\sin \theta + \cos \theta$ is $\sqrt{2}$ (when $\theta = 45^\circ$), which is approximately $1.414$, and the minimum value is $-\sqrt{2}$ (when $\theta = 225^\circ$). The value of $\sin \theta + \cos \theta$ lies in the range $[-\sqrt{2}, \sqrt{2}]$. This range includes values less than or equal to 1.

True

Justification:

We need to determine if the value of $\tan \theta$ increases as $\theta$ increases for $0^\circ \leq \theta < 90^\circ$. This range corresponds to the first quadrant.

Consider the definition of $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

As $\theta$ increases from $0^\circ$ to $90^\circ$ in the first quadrant:

- The value of $\sin \theta$ increases from 0 to 1.

- The value of $\cos \theta$ decreases from 1 to 0.

Since the numerator ($\sin \theta$) is increasing and the denominator ($\cos \theta$) is decreasing (and remaining positive), the ratio $\frac{\sin \theta}{\cos \theta} = \tan \theta$ will increase.

Let's look at some specific values:

$\tan 0^\circ = 0$

$\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$

$\tan 45^\circ = 1$

$\tan 60^\circ = \sqrt{3} \approx 1.732$

As $\theta$ approaches $90^\circ$, $\sin \theta$ approaches 1 and $\cos \theta$ approaches 0, so $\tan \theta$ approaches infinity.

The values $0 < 0.577 < 1 < 1.732 < \ldots$ show that the value of $\tan \theta$ increases as $\theta$ increases from $0^\circ$ towards $90^\circ$.

True

Justification:

We need to consider the rate of increase of $\tan \theta$ compared to $\sin \theta$ as $\theta$ increases, typically in the range $0^\circ \leq \theta < 90^\circ$ (the first quadrant) where both functions are increasing.

The tangent function is defined as $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

For angles $0^\circ < \theta < 90^\circ$, we know that $0 < \cos \theta < 1$.

When we divide $\sin \theta$ by a positive number less than 1, the result ($\tan \theta$) is greater than $\sin \theta$.

So, for $0^\circ < \theta < 90^\circ$, $\tan \theta > \sin \theta$.

Consider how both functions change as $\theta$ approaches $90^\circ$.

As $\theta \to 90^\circ$, $\sin \theta \to 1$. The maximum value $\sin \theta$ reaches in the first quadrant is 1.

As $\theta \to 90^\circ$, $\cos \theta \to 0$ (from the positive side).

As $\theta \to 90^\circ$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$ approaches $\frac{1}{0}$ which tends to infinity ($\infty$).

Since $\tan \theta$ increases from 0 towards $\infty$ as $\theta$ goes from $0^\circ$ to $90^\circ$, while $\sin \theta$ only increases from 0 towards 1, the rate at which $\tan \theta$ increases is much faster than that of $\sin \theta$ in this interval.

False

Justification:

The value of $\sin \theta$ for any real angle $\theta$ always lies in the range $[-1, 1]$.

Thus, the maximum possible value for $\sin \theta$ is $1$ and the minimum value is $-1$.

Now consider the expression $a + \frac{1}{a}$, where $a$ is a positive number ($a > 0$).

For any positive real number $a$, it is a known property that $a + \frac{1}{a} \geq 2$.

This can be shown using the AM-GM inequality (Arithmetic Mean is greater than or equal to Geometric Mean) for positive numbers $a$ and $\frac{1}{a}$:

$\frac{a + \frac{1}{a}}{2} \geq \sqrt{a \times \frac{1}{a}}$

$\frac{a + \frac{1}{a}}{2} \geq \sqrt{1}$

$\frac{a + \frac{1}{a}}{2} \geq 1$

$a + \frac{1}{a} \geq 2$

The equality $a + \frac{1}{a} = 2$ holds only when $a = 1$. For any other positive value of $a$, $a + \frac{1}{a} > 2$.

Since the value of $a + \frac{1}{a}$ is always greater than or equal to 2 for $a > 0$, and the maximum possible value of $\sin \theta$ is 1, $\sin \theta$ can never be equal to $a + \frac{1}{a}$ when $a$ is a positive number.

True

Justification:

We need to check if the statement $\frac{\tan 47^\circ}{\cot 43^\circ} = 1$ is true.

We know the complementary angle identity: $\cot x = \tan (90^\circ - x)$.

Let $x = 43^\circ$. Then $90^\circ - x = 90^\circ - 43^\circ = 47^\circ$.

So, $\cot 43^\circ = \tan (90^\circ - 43^\circ) = \tan 47^\circ$.

Substitute this into the given expression:

$\frac{\tan 47^\circ}{\cot 43^\circ} = \frac{\tan 47^\circ}{\tan 47^\circ}$

Assuming $\tan 47^\circ \neq 0$, which is true for $47^\circ$, we can cancel the term:

$\frac{\tan 47^\circ}{\tan 47^\circ} = 1$

Since the value of the expression is 1, the given statement $\frac{\tan 47^\circ}{\cot 43^\circ} = 1$ is true.

False

Justification:

We need to evaluate the expression $(\cos^2 23^\circ – \sin^2 67^\circ)$ and determine if its value is positive.

We use the complementary angle identity: $\sin x = \cos (90^\circ - x)$.

Notice that $23^\circ + 67^\circ = 90^\circ$. So, $67^\circ = 90^\circ - 23^\circ$.

Using the identity, we can rewrite $\sin 67^\circ$:

$\sin 67^\circ = \sin (90^\circ - 23^\circ) = \cos 23^\circ$.

Now, substitute this into the given expression:

$\cos^2 23^\circ – \sin^2 67^\circ = \cos^2 23^\circ – (\cos 23^\circ)^2$

$\cos^2 23^\circ – \sin^2 67^\circ = \cos^2 23^\circ – \cos^2 23^\circ$

$\cos^2 23^\circ – \sin^2 67^\circ = 0$

The value of the expression is 0. Since 0 is neither positive nor negative, the statement that the value is positive is false.

False

Justification:

We need to determine if the value of the expression $(\sin 80^\circ – \cos 80^\circ)$ is negative.

Consider the angles in the first quadrant ($0^\circ \leq \theta \leq 90^\circ$).

We know that $\sin \theta$ increases from 0 to 1 as $\theta$ increases from $0^\circ$ to $90^\circ$.

We know that $\cos \theta$ decreases from 1 to 0 as $\theta$ increases from $0^\circ$ to $90^\circ$.

The point where $\sin \theta = \cos \theta$ in the first quadrant is $\theta = 45^\circ$.

For angles $0^\circ \leq \theta < 45^\circ$, $\cos \theta > \sin \theta$, so $\sin \theta - \cos \theta$ is negative.

For angles $45^\circ < \theta \leq 90^\circ$, $\sin \theta > \cos \theta$, so $\sin \theta - \cos \theta$ is positive.

For $\theta = 45^\circ$, $\sin \theta - \cos \theta = 0$.

In this question, the angle is $80^\circ$. Since $80^\circ$ is in the first quadrant and $80^\circ > 45^\circ$, we have $\sin 80^\circ > \cos 80^\circ$.

Therefore, the difference $\sin 80^\circ - \cos 80^\circ$ must be positive.

For example, $\sin 80^\circ \approx 0.9848$ and $\cos 80^\circ \approx 0.1736$.

$\sin 80^\circ - \cos 80^\circ \approx 0.9848 - 0.1736 = 0.8112$, which is a positive value.

Since the value of the expression is positive, the statement that the value is negative is false.

False

Justification:

We need to check if the expression $\sqrt{(1 - \cos^2\theta)\sec^2\theta}$ is equal to $\tan\theta$ for all values of $\theta$ where the terms are defined.

Consider the expression under the square root: $(1 - \cos^2\theta)\sec^2\theta$.

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we have $1 - \cos^2\theta = \sin^2\theta$.

So, the expression becomes $\sin^2\theta \cdot \sec^2\theta$.

We know that $\sec\theta = \frac{1}{\cos\theta}$, so $\sec^2\theta = \frac{1}{\cos^2\theta}$.

Substituting this, we get $\sin^2\theta \cdot \frac{1}{\cos^2\theta} = \frac{\sin^2\theta}{\cos^2\theta}$.

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we have $\tan^2\theta = \left(\frac{\sin\theta}{\cos\theta}\right)^2 = \frac{\sin^2\theta}{\cos^2\theta}$.

So, the expression under the square root is $\tan^2\theta$.

Therefore, $\sqrt{(1 - \cos^2\theta)\sec^2\theta} = \sqrt{\tan^2\theta}$.

The square root of a square of a real number is the absolute value of the number:

$\sqrt{\tan^2\theta} = |\tan\theta|$.

The given statement is $\sqrt{(1 - \cos^2\theta)\sec^2\theta} = \tan\theta$, which simplifies to $|\tan\theta| = \tan\theta$.

This equality $|\tan\theta| = \tan\theta$ is only true when $\tan\theta \geq 0$.

$\tan\theta$ is positive in the first and third quadrants ($0^\circ \leq \theta < 90^\circ$ and $180^\circ \leq \theta < 270^\circ$).

However, $\tan\theta$ is negative in the second and fourth quadrants ($90^\circ < \theta < 180^\circ$ and $270^\circ < \theta < 360^\circ$). For example, if $\theta = 135^\circ$, $\tan 135^\circ = -1$, but $\sqrt{(1 - \cos^2 135^\circ)\sec^2 135^\circ} = \sqrt{\tan^2 135^\circ} = \sqrt{(-1)^2} = \sqrt{1} = 1$. Here, $1 \neq -1$.

Since $|\tan\theta|$ is not always equal to $\tan\theta$, the statement is false.

True

Justification:

We are given the equation: $\cos A + \cos^2 A = 1$.

Rearranging the terms in the given equation, we get:

$\cos A = 1 - \cos^2 A$

Using the fundamental trigonometric identity, $\sin^2 \theta + \cos^2 \theta = 1$, we know that $1 - \cos^2 A = \sin^2 A$.

Substituting this into the rearranged equation from the given condition:

$\cos A = \sin^2 A$

Now, consider the expression we need to evaluate: $\sin^2 A + \sin^4 A$.

We can rewrite $\sin^4 A$ as $(\sin^2 A)^2$.

So, the expression is $\sin^2 A + (\sin^2 A)^2$.

Substitute the relationship $\sin^2 A = \cos A$ into this expression:

$\sin^2 A + (\sin^2 A)^2 = (\cos A) + (\cos A)^2 = \cos A + \cos^2 A$

From the initial given condition, we know that $\cos A + \cos^2 A = 1$.

Therefore, $\sin^2 A + \sin^4 A = 1$.

Since the value of the expression $\sin^2 A + \sin^4 A$ is 1, the given statement is true.

False

Justification:

We need to check if the given equation is true for all values of $\theta$ for which the expressions are defined.

Consider the Left Hand Side (LHS) of the equation: $(\tan \theta + 2) (2 \tan \theta + 1)$.

Expand the product:

$(\tan \theta + 2) (2 \tan \theta + 1) = \tan \theta (2 \tan \theta) + \tan \theta (1) + 2 (2 \tan \theta) + 2 (1)$

$= 2 \tan^2 \theta + \tan \theta + 4 \tan \theta + 2$

Combine like terms:

LHS $= 2 \tan^2 \theta + 5 \tan \theta + 2$

Now, consider the Right Hand Side (RHS) of the equation: $5 \tan \theta + \sec^2 \theta$.

We use the fundamental trigonometric identity: $\sec^2 \theta = 1 + \tan^2 \theta$.

Substitute this identity into the RHS:

RHS $= 5 \tan \theta + (1 + \tan^2 \theta)$

Rearrange the terms:

RHS $= \tan^2 \theta + 5 \tan \theta + 1$

Comparing the simplified LHS and RHS:

LHS $= 2 \tan^2 \theta + 5 \tan \theta + 2$

RHS $= \tan^2 \theta + 5 \tan \theta + 1$

The expressions for LHS and RHS are not equal ($2 \tan^2 \theta + 5 \tan \theta + 2 \neq \tan^2 \theta + 5 \tan \theta + 1$). For the equality to hold, we would need $\tan^2 \theta + 1 = 0$, or $\tan^2 \theta = -1$, which is not possible for any real angle $\theta$.

Since the LHS is not equal to the RHS for all valid values of $\theta$, the given statement is false.

False

Justification:

Let the height of the tower be $h$ (which is constant) and the length of the shadow be $s$. Let the angle of elevation of the sun be $\theta$.

In the right-angled triangle formed by the tower, its shadow, and the line of sight from the end of the shadow to the top of the tower, we have:

$\tan \theta = \frac{\text{Height of the tower}}{\text{Length of the shadow}}$

$\tan \theta = \frac{h}{s}$

We are given that the length of the shadow ($s$) is increasing.

The height of the tower ($h$) is constant.

Consider the relationship $\tan \theta = \frac{h}{s}$. If $h$ is constant and $s$ is increasing, the value of the ratio $\frac{h}{s}$ decreases.

So, as the length of the shadow increases, $\tan \theta$ decreases.

For angles in the first quadrant ($0^\circ < \theta < 90^\circ$, which is the relevant range for the sun's elevation when casting a shadow), the tangent function is an increasing function. This means that if $\tan \theta$ decreases, the angle $\theta$ must also decrease.

For example, we know that $\tan 60^\circ = \sqrt{3} \approx 1.732$ and $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$. A smaller tangent value corresponds to a smaller angle.

Therefore, if the length of the shadow of a tower is increasing, the angle of elevation of the sun is decreasing.

The statement that the angle of elevation is increasing is false.

False

Justification:

Let the height of the platform above the lake surface be $h = 3$ metres.

Let the height of the cloud above the lake surface be $H$ metres.

Let the horizontal distance from the observer to the point directly below the cloud on the lake surface be $d$ metres.

The observer is at a point P, $h$ metres above the lake surface. Let Q be the point on the vertical line below the cloud such that PQ is horizontal. The height of Q above the lake surface is equal to the height of P above the lake surface, which is $h$. The horizontal distance PQ is $d$.

The height of the cloud C above the horizontal line PQ is $CQ = H - h$. The angle of elevation of the cloud ($\alpha$) is the angle between PQ and PC. In the right-angled triangle $\triangle$PQC, we have:

$\tan \alpha = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{CQ}{PQ} = \frac{H-h}{d}$

The reflection of the cloud R is formed below the lake surface at a depth equal to the height of the cloud above the surface. So, the distance of the reflection R below the lake surface is $H$.

The point M on the lake surface is directly below the cloud C and directly above the reflection R. The distance from the horizontal line PQ down to the lake surface is $QM = h$. The distance from the lake surface M down to the reflection R is $MR = H$.

The total vertical distance from the horizontal line PQ down to the reflection R is $QR = QM + MR = h + H$. The angle of depression of the reflection ($\beta$) is the angle between the horizontal line through P and PR. In the right-angled triangle $\triangle$PQR, we have:

$\tan \beta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{QR}{PQ} = \frac{h+H}{d}$

Comparing the tangent of the angle of elevation and the angle of depression:

$\tan \alpha = \frac{H-h}{d}$

$\tan \beta = \frac{H+h}{d}$

Given that the platform is 3 metres above the lake, $h = 3$. We assume the cloud is above the observer, so $H > h$. Therefore, $H-h$ is a positive value, and $H+h$ is a positive value. Since $h > 0$, we have $H+h > H-h$.

As the horizontal distance $d$ is the same, we have $\frac{H+h}{d} > \frac{H-h}{d}$, which implies $\tan \beta > \tan \alpha$.

For angles of elevation and depression (which are positive acute angles in this context), the tangent function is strictly increasing. Therefore, if $\tan \beta > \tan \alpha$, it implies $\beta > \alpha$.

The angle of depression of the reflection ($\beta$) is greater than the angle of elevation of the cloud ($\alpha$) when the observer is above the lake surface ($h > 0$). The angles are equal only if the observer is at the surface of the lake ($h=0$), which is not the case here.

Therefore, the statement that the angle of elevation of the cloud is equal to the angle of depression of its reflection is false.

False

Justification:

The range of the sine function is $-1 \leq \sin \theta \leq 1$ for any real angle $\theta$.

Therefore, the range of $2\sin \theta$ is $2 \times (-1) \leq 2\sin \theta \leq 2 \times 1$, which is $-2 \leq 2\sin \theta \leq 2$.

The maximum possible value for $2\sin \theta$ is 2.

Now consider the expression $a + \frac{1}{a}$, where $a$ is a positive number and $a \neq 1$.

For any positive number $a$, the Arithmetic Mean-Geometric Mean (AM-GM) inequality states that $\frac{a + \frac{1}{a}}{2} \geq \sqrt{a \times \frac{1}{a}}$.

$\frac{a + \frac{1}{a}}{2} \geq \sqrt{1}$

$\frac{a + \frac{1}{a}}{2} \geq 1$

$a + \frac{1}{a} \geq 2$

The equality $a + \frac{1}{a} = 2$ holds if and only if $a = \frac{1}{a}$, which means $a^2 = 1$. Since $a$ is positive, this gives $a = 1$.

However, the problem states that $a \neq 1$. Therefore, for a positive number $a$ where $a \neq 1$, the equality $a + \frac{1}{a} = 2$ does not hold. In this case, the strict inequality $a + \frac{1}{a} > 2$ is true.

So, the value of $a + \frac{1}{a}$ is always strictly greater than 2 when $a > 0$ and $a \neq 1$.

Since the maximum value that $2\sin \theta$ can take is 2, and the minimum value that $a + \frac{1}{a}$ takes under the given conditions ($a>0, a \neq 1$) is strictly greater than 2, $2\sin \theta$ can never be equal to $a + \frac{1}{a}$.

False

Justification:

We need to determine if the value of $\cos \theta$ can be equal to the expression $\frac{a^2 + b^2}{2ab}$ under the given conditions that $a$ and $b$ are two distinct numbers ($a \neq b$) and their product $ab > 0$.

The range of the cosine function is $-1 \leq \cos \theta \leq 1$ for any real angle $\theta$. This means that the value of $\cos \theta$ cannot be greater than 1 or less than -1.

Now consider the expression $\frac{a^2 + b^2}{2ab}$ with the conditions $a \neq b$ and $ab > 0$.

Since $ab > 0$, $a$ and $b$ must have the same sign (both positive or both negative).

Consider the property of real numbers that the square of a non-zero number is always positive. Since $a \neq b$, the difference $(a-b)$ is non-zero, so $(a-b)^2 > 0$.

Expanding $(a-b)^2 > 0$, we get:

$a^2 - 2ab + b^2 > 0$

Add $2ab$ to both sides:

$a^2 + b^2 > 2ab$

Now, divide both sides of the inequality by $2ab$. Since $ab > 0$, $2ab$ is a positive number, so the inequality direction does not change.

$\frac{a^2 + b^2}{2ab} > \frac{2ab}{2ab}$

$\frac{a^2 + b^2}{2ab} > 1$

This shows that for any two distinct numbers $a$ and $b$ with $ab > 0$, the value of the expression $\frac{a^2 + b^2}{2ab}$ is always greater than 1.

Since the value of $\cos \theta$ cannot be greater than 1, $\cos \theta$ can never be equal to $\frac{a^2 + b^2}{2ab}$ under the given conditions.

False

Justification:

Let the initial height of the tower be $h$ and the horizontal distance from the observation point to the base of the tower be $x$. Let the initial angle of elevation be $\theta_1$.

From the definition of tangent in a right-angled triangle, we have:

$\tan \theta_1 = \frac{\text{Height of tower}}{\text{Horizontal distance}}$

$\tan \theta_1 = \frac{h}{x}$

We are given that the initial angle of elevation is $30^\circ$. So, $\theta_1 = 30^\circ$.

$\tan 30^\circ = \frac{h}{x}$

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

So, $\frac{1}{\sqrt{3}} = \frac{h}{x}$

This gives us the horizontal distance $x$ in terms of $h$: $x = h\sqrt{3}$.

Now, the height of the tower is doubled. Let the new height be $h' = 2h$. The horizontal distance $x$ remains the same.

Let the new angle of elevation be $\theta_2$.

$\tan \theta_2 = \frac{\text{New height of tower}}{\text{Horizontal distance}}$

$\tan \theta_2 = \frac{h'}{x} = \frac{2h}{x}$

Substitute the value of $x = h\sqrt{3}$ into the equation for $\tan \theta_2$:

$\tan \theta_2 = \frac{2h}{h\sqrt{3}}$

$\tan \theta_2 = \frac{2}{\sqrt{3}}$

According to the statement, if the height is doubled, the angle of elevation will also be doubled. The initial angle was $30^\circ$, so the doubled angle would be $2 \times 30^\circ = 60^\circ$.

Let's check the value of $\tan 60^\circ$.

$\tan 60^\circ = \sqrt{3}$

We found that $\tan \theta_2 = \frac{2}{\sqrt{3}}$ and $\tan 60^\circ = \sqrt{3}$.

Since $\frac{2}{\sqrt{3}} \neq \sqrt{3}$ (because $2 \neq \sqrt{3} \times \sqrt{3} = 3$), $\tan \theta_2 \neq \tan 60^\circ$.

This means the new angle of elevation $\theta_2$ is not $60^\circ$.

Therefore, doubling the height of the tower does not double the angle of elevation from a fixed point.

True

Justification:

Let the original height of the tower be $h$ and the original distance of the point of observation from its foot be $x$. Let the original angle of elevation be $\theta$.

In the right-angled triangle formed by the tower, the horizontal distance, and the line of sight to the top, we have the relationship:

$\tan \theta = \frac{\text{Height of tower}}{\text{Distance from foot}} = \frac{h}{x}$

Now, the height of the tower is increased by 10%. The new height, $h'$, is:

$h' = h + 10\% \text{ of } h = h + 0.10h = 1.10h$

The distance of the point of observation from the foot is also increased by 10%. The new distance, $x'$, is:

$x' = x + 10\% \text{ of } x = x + 0.10x = 1.10x$

Let the new angle of elevation be $\theta'$. The relationship between the new height, new distance, and new angle of elevation is:

$\tan \theta' = \frac{\text{New height of tower}}{\text{New distance from foot}} = \frac{h'}{x'}$

Substitute the expressions for $h'$ and $x'$:

$\tan \theta' = \frac{1.10h}{1.10x}$

Simplify the expression by cancelling the common factor $1.10$:

$\tan \theta' = \frac{\cancel{1.10}}{\cancel{1.10}} \times \frac{h}{x}$

$\tan \theta' = \frac{h}{x}$

We see that the new tangent value, $\tan \theta'$, is equal to the original tangent value, $\tan \theta$ (since $\tan \theta = \frac{h}{x}$).

$\tan \theta' = \tan \theta$

Since the angle of elevation is typically considered in the range $0^\circ < \theta < 90^\circ$, where the tangent function is strictly increasing and unique for each angle, having equal tangent values implies that the angles themselves are equal.

Therefore, $\theta' = \theta$.

The angle of elevation of the top of the tower remains unchanged.

Solution:

We need to prove the identity: $\sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta = 1$.

We start with the fundamental trigonometric identity:

$\sin^2 \theta + \cos^2 \theta = 1$

Cube both sides of this identity:

$(\sin^2 \theta + \cos^2 \theta)^3 = 1^3$

$(\sin^2 \theta + \cos^2 \theta)^3 = 1$

We use the algebraic identity for the cube of a sum: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$.

Let $a = \sin^2 \theta$ and $b = \cos^2 \theta$.

Substitute these into the algebraic identity:

$(\sin^2 \theta)^3 + (\cos^2 \theta)^3 + 3(\sin^2 \theta)(\cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = 1$

Simplify the terms:

$\sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1$

We know that $\sin^2 \theta + \cos^2 \theta = 1$. Substitute this back into the equation:

$\sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta (1) = 1$

$\sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta = 1$

This is the identity we were asked to prove.

Hence, the identity is proved.

Solution:

We need to prove the identity: $(\sin^4 \theta – \cos^4 \theta + 1) \text{cosec}^2 \theta = 2$.

Consider the Left Hand Side (LHS):

LHS $= (\sin^4 \theta – \cos^4 \theta + 1) \text{cosec}^2 \theta$

We can factor the term $(\sin^4 \theta – \cos^4 \theta)$ as a difference of squares:

$\sin^4 \theta – \cos^4 \theta = (\sin^2 \theta)^2 – (\cos^2 \theta)^2 = (\sin^2 \theta – \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we substitute this into the factored expression:

$\sin^4 \theta – \cos^4 \theta = (\sin^2 \theta – \cos^2 \theta)(1) = \sin^2 \theta – \cos^2 \theta$

Now substitute this back into the LHS expression:

LHS $= (\sin^2 \theta – \cos^2 \theta + 1) \text{cosec}^2 \theta$

Rearrange the terms inside the parenthesis:

LHS $= (\sin^2 \theta + (1 – \cos^2 \theta)) \text{cosec}^2 \theta$

Using the fundamental trigonometric identity again, $1 – \cos^2 \theta = \sin^2 \theta$, we substitute this into the parenthesis:

LHS $= (\sin^2 \theta + \sin^2 \theta) \text{cosec}^2 \theta$

Combine the terms inside the parenthesis:

LHS $= (2\sin^2 \theta) \text{cosec}^2 \theta$

Using the reciprocal identity $\text{cosec} \theta = \frac{1}{\sin \theta}$, which means $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ (provided $\sin \theta \neq 0$), substitute this into the expression:

LHS $= 2\sin^2 \theta \times \frac{1}{\sin^2 \theta}$

Assuming $\sin^2 \theta \neq 0$, we can cancel the term $\sin^2 \theta$:

LHS $= 2 \times 1$

LHS $= 2$

The Right Hand Side (RHS) is 2.

Since LHS = RHS, the identity is proved.

Solution:

Given:

$\alpha + \beta = 90^\circ$

To Show:

$\sqrt{\cos\alpha\; \text{cosec}\;\beta - \cos\alpha \sin\beta} = \sin\alpha$

From the given condition $\alpha + \beta = 90^\circ$, we can write $\beta = 90^\circ - \alpha$.

Using complementary angle identities, we have:

$\text{cosec}\;\beta = \text{cosec}\;(90^\circ - \alpha) = \sec \alpha$

$\sin\beta = \sin(90^\circ - \alpha) = \cos \alpha$

Consider the expression inside the square root on the Left Hand Side (LHS):

Expression $= \cos\alpha\; \text{cosec}\;\beta - \cos\alpha \sin\beta$

Substitute the complementary angle identities into the expression:

Expression $= \cos\alpha\; (\sec \alpha) - \cos\alpha (\cos \alpha)$

Expression $= \cos\alpha\; \sec \alpha - \cos^2 \alpha$

Using the reciprocal identity $\sec \alpha = \frac{1}{\cos \alpha}$ (assuming $\cos \alpha \neq 0$), substitute this into the expression:

Expression $= \cos\alpha\; \left(\frac{1}{\cos \alpha}\right) - \cos^2 \alpha$

Expression $= 1 - \cos^2 \alpha$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we know that $1 - \cos^2 \alpha = \sin^2 \alpha$.

So, the expression inside the square root simplifies to $\sin^2 \alpha$.

Now, take the square root of this expression:

LHS $= \sqrt{\cos\alpha\; \text{cosec}\;\beta - \cos\alpha \sin\beta} = \sqrt{\sin^2 \alpha}$

The square root of a square of a real number is the absolute value of the number:

LHS $= |\sin \alpha|$

The Right Hand Side (RHS) of the identity to be proved is $\sin \alpha$.

So, we need to show that $|\sin \alpha| = \sin \alpha$. This is true if and only if $\sin \alpha \geq 0$.

In the context of trigonometric identities without specified angle ranges, it is often assumed that the angles are such that the principal values or positive values are considered. Given the form of the identity $\sqrt{\dots} = \sin\alpha$, it implies that $\sin\alpha$ must be non-negative.

Assuming that $\alpha$ is in a range where $\sin \alpha \geq 0$ (e.g., $0^\circ \leq \alpha \leq 180^\circ$, provided the other terms are defined):

LHS $= |\sin \alpha| = \sin \alpha$

RHS $= \sin \alpha$

LHS = RHS

Hence, the identity is shown to be true under the assumption that $\sin \alpha \geq 0$.

Solution:

Given:

$\sin \theta + \cos \theta = \sqrt{3}$

To Prove:

$\tan \theta + \cot \theta = 1$

Proof:

Start with the given equation:

$\sin \theta + \cos \theta = \sqrt{3}$

Square both sides of the equation:

$(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$

Expand the left side using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 3$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, substitute this into the equation:

$1 + 2\sin \theta \cos \theta = 3$

Subtract 1 from both sides:

$2\sin \theta \cos \theta = 3 - 1$

$2\sin \theta \cos \theta = 2$

Divide both sides by 2:

$\sin \theta \cos \theta = 1$

Now, consider the Left Hand Side (LHS) of the identity we want to prove: $\tan \theta + \cot \theta$.

Rewrite $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$:

LHS $= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$

Combine the fractions by finding a common denominator, which is $\sin \theta \cos \theta$:

LHS $= \frac{\sin \theta \times \sin \theta + \cos \theta \times \cos \theta}{\cos \theta \sin \theta}$

LHS $= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ in the numerator:

LHS $= \frac{1}{\sin \theta \cos \theta}$

Substitute the value $\sin \theta \cos \theta = 1$ (derived earlier) into this expression:

LHS $= \frac{1}{1}$

LHS $= 1$

The Right Hand Side (RHS) of the identity to be proved is 1.

Since LHS = RHS, the identity is proved.

Solution:

We need to prove the identity: $\frac{\sin\; \theta}{1 \;+\; \cos\; \theta}$ + $\frac{1 \;+\; \cos\; \theta}{\sin\; \theta}$ = 2 $\text{cosec}\;\theta$.

Consider the Left Hand Side (LHS):

LHS $= \frac{\sin\; \theta}{1 \;+\; \cos\; \theta}$ + $\frac{1 \;+\; \cos\; \theta}{\sin\; \theta}$

Combine the fractions by finding a common denominator, which is $(\sin\; \theta)(1 \;+\; \cos\; \theta)$:

LHS $= \frac{(\sin\; \theta)(\sin\; \theta) \;+\; (1 \;+\; \cos\; \theta)(1 \;+\; \cos\; \theta)}{(\sin\; \theta)(1 \;+\; \cos\; \theta)}$

LHS $= \frac{\sin^2\; \theta \;+\; (1 \;+\; \cos\; \theta)^2}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

Expand the term $(1 \;+\; \cos\; \theta)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$(1 \;+\; \cos\; \theta)^2 = 1^2 + 2(1)(\cos\; \theta) + (\cos\; \theta)^2 = 1 + 2\cos\; \theta + \cos^2\; \theta$

Substitute the expanded term back into the numerator of the LHS:

LHS $= \frac{\sin^2\; \theta \;+\; 1 + 2\cos\; \theta + \cos^2\; \theta}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

Group the terms $\sin^2\; \theta + \cos^2\; \theta$ in the numerator:

LHS $= \frac{(\sin^2\; \theta + \cos^2\; \theta) \;+\; 1 + 2\cos\; \theta}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

Using the fundamental trigonometric identity $\sin^2\; \theta + \cos^2\; \theta = 1$, substitute 1 into the numerator:

LHS $= \frac{1 \;+\; 1 + 2\cos\; \theta}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

LHS $= \frac{2 + 2\cos\; \theta}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

Factor out the common factor 2 from the numerator:

LHS $= \frac{2(1 \;+\; \cos\; \theta)}{\sin\; \theta(1 \;+\; \cos\; \theta)}$

Cancel out the common term $(1 \;+\; \cos\; \theta)$ from the numerator and the denominator (assuming $1 + \cos \theta \neq 0$, which is true if $\theta \neq (2n+1)\pi$ for integer $n$):

LHS $= \frac{2}{\sin\; \theta}$

Using the reciprocal identity $\text{cosec}\;\theta = \frac{1}{\sin\; \theta}$, substitute this into the expression:

LHS $= 2 \times \frac{1}{\sin\; \theta}$

LHS $= 2 \text{cosec}\;\theta$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Solution:

We need to prove the identity: $\frac{\tan\; A}{1 \;+\; \sec\;A}$ - $\frac{\tan\; A}{1 \;-\; \sec\;A}$ = 2 $\text{cosec}\;$A.

Consider the Left Hand Side (LHS):

LHS $= \frac{\tan\; A}{1 \;+\; \sec\;A}$ - $\frac{\tan\; A}{1 \;-\; \sec\;A}$

Combine the fractions by finding a common denominator, which is $(1 \;+\; \sec\;A)(1 \;-\; \sec\;A)$:

LHS $= \frac{\tan\; A (1 \;-\; \sec\;A) \;-\; \tan\; A (1 \;+\; \sec\;A)}{(1 \;+\; \sec\;A)(1 \;-\; \sec\;A)}$

Expand the numerator:

Numerator $= \tan\; A - \tan\; A \sec\; A - (\tan\; A + \tan\; A \sec\; A)$

Numerator $= \tan\; A - \tan\; A \sec\; A - \tan\; A - \tan\; A \sec\; A$

Numerator $= (\tan\; A - \tan\; A) + (-\tan\; A \sec\; A - \tan\; A \sec\; A)$

Numerator $= 0 - 2 \tan\; A \sec\; A$

Numerator $= -2 \tan\; A \sec\; A$

Expand the denominator using the difference of squares formula $(x+y)(x-y) = x^2 - y^2$:

Denominator $= (1 \;+\; \sec\;A)(1 \;-\; \sec\;A) = 1^2 - \sec^2\; A = 1 - \sec^2\; A$

Substitute the expanded numerator and denominator back into the LHS expression:

LHS $= \frac{-2 \tan\; A \sec\; A}{1 - \sec^2\; A}$

Using the trigonometric identity $1 + \tan^2\; \theta = \sec^2\; \theta$, we can write $1 - \sec^2\; A = -\tan^2\; A$.

Substitute this into the denominator:

LHS $= \frac{-2 \tan\; A \sec\; A}{-\tan^2\; A}$

Cancel the negative signs and simplify the expression (assuming $\tan\; A \neq 0$):

LHS $= \frac{2 \cancel{\tan\; A} \sec\; A}{\tan^{\cancel{2}}\; A} = \frac{2 \sec\; A}{\tan\; A}$

Rewrite $\sec\; A$ and $\tan\; A$ in terms of $\sin\; A$ and $\cos\; A$:

$\sec\; A = \frac{1}{\cos\; A}$

$\tan\; A = \frac{\sin\; A}{\cos\; A}$

Substitute these into the LHS expression:

LHS $= \frac{2 \times \frac{1}{\cos\; A}}{\frac{\sin\; A}{\cos\; A}}$

Simplify the complex fraction (assuming $\cos\; A \neq 0$):

LHS $= \frac{2}{\cos\; A} \times \frac{\cos\; A}{\sin\; A} = \frac{2 \cancel{\cos\; A}}{\cancel{\cos\; A} \sin\; A} = \frac{2}{\sin\; A}$

Using the reciprocal identity $\text{cosec}\; A = \frac{1}{\sin\; A}$ (assuming $\sin\; A \neq 0$):

LHS $= 2 \times \frac{1}{\sin\; A} = 2 \text{cosec}\; A$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved for all values of A where the expressions are defined.

Solution:

Given:

$\tan A = \frac{3}{4}$

To Prove:

$\sin A \cos A = \frac{12}{25}$

Proof:

We are given that $\tan A = \frac{3}{4}$. We know that $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}}$ in a right-angled triangle.

Consider a right-angled triangle where the angle is A.

Let the side opposite to angle A be $3k$ and the adjacent side be $4k$, for some positive constant $k$.

Using the Pythagoras theorem, the hypotenuse ($h$) is given by:

$h^2 = (\text{Opposite side})^2 + (\text{Adjacent side})^2$

$h^2 = (3k)^2 + (4k)^2$

$h^2 = 9k^2 + 16k^2$

$h^2 = 25k^2$

$h = \sqrt{25k^2} = 5k$ (Since length must be positive, we take the positive root)

Now we can find the values of $\sin A$ and $\cos A$ using the side lengths:

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}$

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}$

Now, calculate the product $\sin A \cos A$:

$\sin A \cos A = \left(\frac{3}{5}\right) \times \left(\frac{4}{5}\right)$

$\sin A \cos A = \frac{3 \times 4}{5 \times 5} = \frac{12}{25}$

Thus, we have shown that if $\tan A = \frac{3}{4}$, then $\sin A \cos A = \frac{12}{25}$.

Note: If angle A were in the third quadrant where $\tan A > 0$, $\sin A$ and $\cos A$ would both be negative ($\sin A = -\frac{3}{5}$ and $\cos A = -\frac{4}{5}$). However, their product would still be $(-\frac{3}{5}) \times (-\frac{4}{5}) = \frac{12}{25}$.

Solution:

We need to prove the identity: $(\sin \alpha + \cos \alpha) (\tan \alpha + \cot \alpha) = \sec \alpha + \text{cosec} \ \alpha$.

Consider the Left Hand Side (LHS):

LHS $= (\sin \alpha + \cos \alpha) (\tan \alpha + \cot \alpha)$

Rewrite $\tan \alpha$ and $\cot \alpha$ in terms of $\sin \alpha$ and $\cos \alpha$:

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$

$\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$

Substitute these into the LHS expression:

LHS $= (\sin \alpha + \cos \alpha) \left(\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}\right)$

Combine the terms in the second parenthesis by finding a common denominator, which is $\sin \alpha \cos \alpha$:

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}$

Using the fundamental trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$ in the numerator:

$\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} = \frac{1}{\sin \alpha \cos \alpha}$

Substitute this back into the LHS expression:

LHS $= (\sin \alpha + \cos \alpha) \left(\frac{1}{\sin \alpha \cos \alpha}\right)$

Distribute the term $(\sin \alpha + \cos \alpha)$:

LHS $= \frac{\sin \alpha}{\sin \alpha \cos \alpha} + \frac{\cos \alpha}{\sin \alpha \cos \alpha}$

Simplify the terms by cancelling common factors (assuming $\sin \alpha \neq 0$ and $\cos \alpha \neq 0$):

LHS $= \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha}$

Using the reciprocal identities $\sec \alpha = \frac{1}{\cos \alpha}$ and $\text{cosec} \alpha = \frac{1}{\sin \alpha}$ (assuming $\sin \alpha \neq 0$ and $\cos \alpha \neq 0$):

LHS $= \sec \alpha + \text{cosec} \ \alpha$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved for all values of $\alpha$ where the expressions are defined.

Solution:

We need to prove the given equation by evaluating both sides.

First, let's find the values of the trigonometric terms involved:

$\cot 30^\circ = \sqrt{3}$

$\tan 60^\circ = \sqrt{3}$

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

Consider the Left Hand Side (LHS):

LHS $= (\sqrt{3} + 1)(3 – \cot 30^\circ)$

Substitute the value of $\cot 30^\circ$:

LHS $= (\sqrt{3} + 1)(3 – \sqrt{3})$

Expand the product using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$, with $a=3$ and $b=\sqrt{3}$, or by direct multiplication:

LHS $= 3(\sqrt{3}) - (\sqrt{3})(\sqrt{3}) + 1(3) - 1(\sqrt{3})$

LHS $= 3\sqrt{3} - 3 + 3 - \sqrt{3}$

LHS $= (3\sqrt{3} - \sqrt{3}) + (-3 + 3)$

LHS $= 2\sqrt{3} + 0$

LHS $= 2\sqrt{3}$

Now, consider the Right Hand Side (RHS):

RHS $= \tan^3 60^\circ – 2 \sin 60^\circ$

Substitute the values of $\tan 60^\circ$ and $\sin 60^\circ$:

RHS $= (\sqrt{3})^3 – 2 \left(\frac{\sqrt{3}}{2}\right)$

Calculate the terms:

$(\sqrt{3})^3 = \sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3 \times \sqrt{3} = 3\sqrt{3}$

$2 \left(\frac{\sqrt{3}}{2}\right) = \cancel{2} \times \frac{\sqrt{3}}{\cancel{2}} = \sqrt{3}$

Substitute these values back into the RHS expression:

RHS $= 3\sqrt{3} – \sqrt{3}$

RHS $= 2\sqrt{3}$

Since LHS $= 2\sqrt{3}$ and RHS $= 2\sqrt{3}$, we have LHS = RHS.

Hence, the given equation is proved.

Solution:

We need to prove the identity: $1 + \frac{\cot^2\alpha}{1 \;+\; \text{cosec}\;\alpha} = \text{cosec}\;\alpha$.

Consider the Left Hand Side (LHS):

LHS $= 1 + \frac{\cot^2\alpha}{1 \;+\; \text{cosec}\;\alpha}$

We use the trigonometric identity: $\cot^2 \alpha + 1 = \text{cosec}^2 \alpha$.

From this identity, we can write $\cot^2 \alpha = \text{cosec}^2 \alpha - 1$.

Substitute this expression for $\cot^2 \alpha$ into the LHS:

LHS $= 1 + \frac{\text{cosec}^2\alpha \;-\; 1}{1 \;+\; \text{cosec}\;\alpha}$

Factor the numerator $(\text{cosec}^2\alpha \;-\; 1)$ as a difference of squares using the formula $x^2 - y^2 = (x-y)(x+y)$, with $x = \text{cosec}\;\alpha$ and $y = 1$:

$\text{cosec}^2\alpha \;-\; 1 = (\text{cosec}\;\alpha \;-\; 1)(\text{cosec}\;\alpha \;+\; 1)$

Substitute the factored numerator back into the LHS expression:

LHS $= 1 + \frac{(\text{cosec}\;\alpha \;-\; 1)(\text{cosec}\;\alpha \;+\; 1)}{1 \;+\; \text{cosec}\;\alpha}$

Cancel out the common term $(1 \;+\; \text{cosec}\;\alpha)$ from the numerator and the denominator (assuming $1 + \text{cosec}\;\alpha \neq 0$, which means $\text{cosec}\;\alpha \neq -1$, or $\sin\alpha \neq -1$, i.e., $\alpha \neq \frac{3\pi}{2} + 2n\pi$ for integer $n$):

LHS $= 1 + (\text{cosec}\;\alpha \;-\; 1)$

Simplify the expression:

LHS $= 1 + \text{cosec}\;\alpha - 1$

LHS $= \text{cosec}\;\alpha$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved for all values of $\alpha$ where the expressions are defined and $1 + \text{cosec}\;\alpha \neq 0$.

Solution:

We need to prove the identity: $\tan \theta + \tan (90^\circ – \theta) = \sec \theta \sec (90^\circ – \theta)$.

We use the complementary angle identities:

$\tan (90^\circ - \theta) = \cot \theta$

$\sec (90^\circ - \theta) = \text{cosec} \theta$

Consider the Left Hand Side (LHS):

LHS $= \tan \theta + \tan (90^\circ – \theta)$

Substitute the complementary angle identity for $\tan (90^\circ – \theta)$:

LHS $= \tan \theta + \cot \theta$

Rewrite $\tan \theta$ and $\cot \theta$ in terms of $\sin \theta$ and $\cos \theta$:

LHS $= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$

Combine the terms by finding a common denominator, which is $\sin \theta \cos \theta$ (assuming $\sin \theta \neq 0$ and $\cos \theta \neq 0$):

LHS $= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ in the numerator:

LHS $= \frac{1}{\sin \theta \cos \theta}$

Now, consider the Right Hand Side (RHS):

RHS $= \sec \theta \sec (90^\circ – \theta)$

Substitute the complementary angle identity for $\sec (90^\circ – \theta)$:

RHS $= \sec \theta \ \text{cosec} \theta$

Rewrite $\sec \theta$ and $\text{cosec} \theta$ in terms of $\sin \theta$ and $\cos \theta$:

RHS $= \frac{1}{\cos \theta} \times \frac{1}{\sin \theta}$

RHS $= \frac{1}{\cos \theta \sin \theta} = \frac{1}{\sin \theta \cos \theta}$

Comparing the simplified LHS and RHS:

LHS $= \frac{1}{\sin \theta \cos \theta}$

RHS $= \frac{1}{\sin \theta \cos \theta}$

Since LHS = RHS, the identity is proved for all values of $\theta$ where the expressions are defined (i.e., $\sin \theta \neq 0$, $\cos \theta \neq 0$, $\tan \theta$ and $\cot \theta$ are defined). This means $\theta \neq n\frac{\pi}{2}$ for integer $n$.

Solution:

Given:

Height of the pole = $h$ metres

Length of the shadow = $\sqrt{3}h$ metres

To Find:

The angle of elevation of the sun.

Let the height of the pole be represented by the vertical side AB and the length of the shadow by the horizontal side BC of a right-angled triangle ABC, where $\angle$B is $90^\circ$. The Sun's angle of elevation is the angle formed at the end of the shadow on the ground to the top of the pole, which is $\angle$BCA. Let this angle be $\theta$.

In the right-angled triangle ABC:

Opposite side to $\theta$ (Height of pole AB) = $h$ metres

Adjacent side to $\theta$ (Length of shadow BC) = $\sqrt{3}h$ metres

We use the tangent trigonometric ratio:

$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$

$\tan \theta = \frac{AB}{BC}$

Substitute the given values:

$\tan \theta = \frac{h}{\sqrt{3}h}$

Simplify the expression:

$\tan \theta = \frac{\cancel{h}}{\sqrt{3}\cancel{h}}$

$\tan \theta = \frac{1}{\sqrt{3}}$

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Comparing the values, we get:

$\tan \theta = \tan 30^\circ$

Since $\theta$ is an angle of elevation (in the range $0^\circ < \theta < 90^\circ$), we have:

$\theta = 30^\circ$

The angle of elevation of the sun is $\mathbf{30^\circ}$.

Solution:

Given:

$\sqrt{3} \tan \theta = 1$

To Find:

The value of $\sin^2 \theta – \cos^2 \theta$.

From the given equation, find the value of $\tan \theta$:

$\tan \theta = \frac{1}{\sqrt{3}}$

We know that for an acute angle $\theta$, $\tan \theta = \frac{1}{\sqrt{3}}$ corresponds to $\theta = 30^\circ$.

Now, find the values of $\sin \theta$ and $\cos \theta$ for $\theta = 30^\circ$:

$\sin 30^\circ = \frac{1}{2}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

Substitute these values into the expression $\sin^2 \theta – \cos^2 \theta$:

$\sin^2 30^\circ – \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2$

Calculate the squares:

$\left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}$

$\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{2^2} = \frac{3}{4}$

Substitute the squared values back into the expression:

$\sin^2 \theta – \cos^2 \theta = \frac{1}{4} – \frac{3}{4}$

$\sin^2 \theta – \cos^2 \theta = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$

The value of $\sin^2 \theta – \cos^2 \theta$ is $\mathbf{-\frac{1}{2}}$.

Alternate Solution (without finding $\theta$):

Given $\tan \theta = \frac{1}{\sqrt{3}}$.

We need to find $\sin^2 \theta – \cos^2 \theta$. We know that $\sin^2 \theta + \cos^2 \theta = 1$.

Also, we can express $\sin^2 \theta$ and $\cos^2 \theta$ in terms of $\tan^2 \theta$.

Divide the identity $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):

$\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$

$\tan^2 \theta + 1 = \sec^2 \theta$

Since $\sec \theta = \frac{1}{\cos \theta}$, $\sec^2 \theta = \frac{1}{\cos^2 \theta}$.

So, $\cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{1}{1 + \tan^2 \theta}$.

From $\sin^2 \theta + \cos^2 \theta = 1$, we have $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{1 + \tan^2 \theta} = \frac{(1 + \tan^2 \theta) - 1}{1 + \tan^2 \theta} = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$.

Now, substitute $\tan \theta = \frac{1}{\sqrt{3}}$ (so $\tan^2 \theta = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$) into the expressions for $\sin^2 \theta$ and $\cos^2 \theta$:

$\sin^2 \theta = \frac{\frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{3+1}{3}} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$

$\cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = 1 \times \frac{3}{4} = \frac{3}{4}$

Now, calculate $\sin^2 \theta – \cos^2 \theta$:

$\sin^2 \theta – \cos^2 \theta = \frac{1}{4} – \frac{3}{4} = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$

Both methods yield the same result.

Solution:

Given:

Length of the ladder = $15$ metres.

Angle the ladder makes with the wall = $60^\circ$.

To Find:

The height of the wall.

Let the vertical wall be represented by AB, the ground by BC, and the ladder by AC. This forms a right-angled triangle ABC, where $\angle$B is the right angle ($90^\circ$).

The length of the ladder is AC = $15$ m.

The angle the ladder makes with the wall is $\angle$BAC = $60^\circ$.

The height of the wall is AB (which we need to find).

In the right-angled triangle ABC, the side AB is adjacent to the angle $\angle$BAC, and AC is the hypotenuse.

We use the cosine trigonometric ratio:

$\cos (\angle BAC) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\cos 60^\circ = \frac{AB}{AC}$

Substitute the given values:

$\cos 60^\circ = \frac{AB}{15}$

We know the standard value of $\cos 60^\circ$ is $\frac{1}{2}$.

$\frac{1}{2} = \frac{AB}{15}$

Solve for AB:

$AB = 15 \times \frac{1}{2}$

$AB = 7.5$

The height of the wall is $7.5$ metres.

Alternate Solution:

In the right-angled triangle ABC, the sum of the two acute angles is $90^\circ$.

$\angle$BAC + $\angle$BCA = $90^\circ$

$\angle$BCA = $90^\circ - 60^\circ = 30^\circ$.

This is the angle of elevation of the top of the wall from the foot of the ladder.

In the right-angled triangle ABC, the side AB is opposite to the angle $\angle$BCA, and AC is the hypotenuse.

We use the sine trigonometric ratio:

$\sin (\angle BCA) = \frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\sin 30^\circ = \frac{AB}{AC}$

Substitute the given values:

$\sin 30^\circ = \frac{AB}{15}$

We know the standard value of $\sin 30^\circ$ is $\frac{1}{2}$.

$\frac{1}{2} = \frac{AB}{15}$

Solve for AB:

$AB = 15 \times \frac{1}{2}$

$AB = 7.5$

The height of the wall is $7.5$ metres.

Solution:

We need to simplify the expression: $(1 + \tan^2 \theta) (1 – \sin \theta) (1 + \sin \theta)$.

Consider the terms $(1 – \sin \theta) (1 + \sin \theta)$. This is in the form of a difference of squares, $(a-b)(a+b) = a^2 - b^2$, where $a=1$ and $b=\sin \theta$.

So, $(1 – \sin \theta) (1 + \sin \theta) = 1^2 – \sin^2 \theta = 1 – \sin^2 \theta$.

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we can rewrite $1 – \sin^2 \theta$ as $\cos^2 \theta$.

So, $(1 – \sin \theta) (1 + \sin \theta) = \cos^2 \theta$.

Now the original expression becomes $(1 + \tan^2 \theta) (\cos^2 \theta)$.

We use another fundamental trigonometric identity: $1 + \tan^2 \theta = \sec^2 \theta$.

Substitute this into the expression:

The expression is now $(\sec^2 \theta) (\cos^2 \theta)$.

Using the reciprocal identity $\sec \theta = \frac{1}{\cos \theta}$, we have $\sec^2 \theta = \frac{1}{\cos^2 \theta}$.

Substitute this into the expression (assuming $\cos \theta \neq 0$):

The expression becomes $\left(\frac{1}{\cos^2 \theta}\right) (\cos^2 \theta)$.

Cancel the common term $\cos^2 \theta$ (assuming $\cos^2 \theta \neq 0$):

$\frac{1}{\cancel{\cos^2 \theta}} \times \cancel{\cos^2 \theta} = 1$

The simplified value of the expression is $\mathbf{1}$.

Solution:

Given:

$2\sin^2 \theta – \cos^2 \theta = 2$

To Find:

The value of $\theta$.

We use the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$.

From this identity, we can express $\cos^2 \theta$ in terms of $\sin^2 \theta$:

$\cos^2 \theta = 1 - \sin^2 \theta$

Substitute this expression for $\cos^2 \theta$ into the given equation:

$2\sin^2 \theta – (1 - \sin^2 \theta) = 2$

Simplify and solve for $\sin^2 \theta$:

$2\sin^2 \theta – 1 + \sin^2 \theta = 2$

$3\sin^2 \theta – 1 = 2$

$3\sin^2 \theta = 2 + 1$

$3\sin^2 \theta = 3$

$\sin^2 \theta = \frac{3}{3}$

$\sin^2 \theta = 1$

Take the square root of both sides:

$\sin \theta = \pm \sqrt{1}$

$\sin \theta = \pm 1$

For the standard angles, $\sin \theta = 1$ when $\theta = 90^\circ$.

Let's check if this value of $\theta$ satisfies the original equation. If $\theta = 90^\circ$, then $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$.

Substitute these values into the given equation:

$2(\sin 90^\circ)^2 – (\cos 90^\circ)^2 = 2(1)^2 – (0)^2 = 2(1) – 0 = 2 – 0 = 2$.

This matches the RHS of the given equation.

Consider the case $\sin \theta = -1$. This occurs when $\theta = 270^\circ$ (or $-\frac{\pi}{2}$). For this angle, $\cos 270^\circ = 0$.

$2(\sin 270^\circ)^2 – (\cos 270^\circ)^2 = 2(-1)^2 – (0)^2 = 2(1) – 0 = 2 – 0 = 2$.

This also matches the RHS.

However, trigonometric questions at this level often assume acute angles or angles within $0^\circ$ to $180^\circ$ unless specified. If the context implies the principal value, then $\theta = 90^\circ$ is the most common answer.

The general solution for $\sin \theta = 1$ is $\theta = 90^\circ + 360^\circ n$ for integer $n$.

The general solution for $\sin \theta = -1$ is $\theta = 270^\circ + 360^\circ n$ for integer $n$.

These can be combined into $\theta = 90^\circ + 180^\circ n$ for integer $n$ (angles where the y-coordinate on the unit circle is $\pm 1$).

For a single value, the principal value or the value in the range $0^\circ \leq \theta < 360^\circ$ is usually expected. Given the options are typically single angles, $\theta = 90^\circ$ is the likely expected answer if an acute or specific angle is implied.

Assuming $\theta$ is in the range $[0^\circ, 360^\circ)$, the values are $90^\circ$ and $270^\circ$. If only one value is expected, and typically in standard problems, it refers to an angle where the functions are defined, and often an acute angle or within the $0-180^\circ$ range. However, $\theta=90^\circ$ makes sense as $\cos 90^\circ=0$.

Let's consider if there's any other way. If we expressed $\sin^2 \theta$ in terms of $\cos^2 \theta$: $\sin^2 \theta = 1 - \cos^2 \theta$.

$2(1 - \cos^2 \theta) – \cos^2 \theta = 2$

$2 - 2\cos^2 \theta – \cos^2 \theta = 2$

$2 - 3\cos^2 \theta = 2$

$-3\cos^2 \theta = 2 - 2$

$-3\cos^2 \theta = 0$

$\cos^2 \theta = 0$

$\cos \theta = 0$

For standard angles, $\cos \theta = 0$ when $\theta = 90^\circ$ or $\theta = 270^\circ$, etc.

In the range $[0^\circ, 360^\circ)$, $\theta = 90^\circ$ or $\theta = 270^\circ$.

Assuming the context implies a specific angle, $\theta = 90^\circ$ is a principal solution.

The value of $\theta$ is $\mathbf{90^\circ}$ (or $\frac{\pi}{2}$ radians) or $270^\circ$ (or $\frac{3\pi}{2}$ radians), and their co-terminal angles.

Solution:

We need to show that $\frac{\cos^2\; (45^\circ \;+ \;\theta) \;+\; \cos^2\; (45^\circ \;- \;\theta)}{\tan \; (60^\circ \;+ \;\theta)\; \tan \; (30^\circ \;- \;\theta)} = 1$.

Consider the numerator of the fraction:

Numerator $= \cos^2\; (45^\circ \;+ \;\theta) \;+\; \cos^2\; (45^\circ \;- \;\theta)$

We use the complementary angle identity: $\cos x = \sin (90^\circ - x)$.

Let $x = 45^\circ + \theta$. Then $90^\circ - x = 90^\circ - (45^\circ + \theta) = 90^\circ - 45^\circ - \theta = 45^\circ - \theta$.

So, $\cos (45^\circ + \theta) = \sin (90^\circ - (45^\circ + \theta)) = \sin (45^\circ - \theta)$.

Therefore, $\cos^2 (45^\circ + \theta) = (\sin (45^\circ - \theta))^2 = \sin^2 (45^\circ - \theta)$.

Substitute this into the numerator expression:

Numerator $= \sin^2\; (45^\circ \;- \;\theta) \;+\; \cos^2\; (45^\circ \;- \;\theta)$

Using the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$, with $A = 45^\circ - \theta$, the numerator simplifies to 1.

Numerator $= 1$

Now, consider the denominator of the fraction:

Denominator $= \tan \; (60^\circ \;+ \;\theta)\; \tan \; (30^\circ \;- \;\theta)$

We use the complementary angle identity: $\tan x = \cot (90^\circ - x)$.

Let $x = 60^\circ + \theta$. Then $90^\circ - x = 90^\circ - (60^\circ + \theta) = 90^\circ - 60^\circ - \theta = 30^\circ - \theta$.

So, $\tan (60^\circ + \theta) = \cot (90^\circ - (60^\circ + \theta)) = \cot (30^\circ - \theta)$.

Substitute this into the denominator expression:

Denominator $= \cot \; (30^\circ \;- \;\theta)\; \tan \; (30^\circ \;- \;\theta)$

Using the reciprocal identity $\cot A = \frac{1}{\tan A}$ (assuming $\tan (30^\circ - \theta) \neq 0$ and $\cot (30^\circ - \theta) \neq 0$), we have $\cot A \tan A = 1$.

Let $A = 30^\circ - \theta$.

Denominator $= \cot \; (30^\circ \;- \;\theta)\; \tan \; (30^\circ \;- \;\theta) = 1$

Now, substitute the simplified numerator and denominator back into the original fraction:

$\frac{\text{Numerator}}{\text{Denominator}} = \frac{1}{1} = 1$

The value of the expression is 1.

Since the value is 1, the identity is shown to be true for all values of $\theta$ for which the expressions are defined.

Solution:

Given:

Height of the observer = $1.5$ metres.

Distance of the observer from the tower = $20.5$ metres.

Height of the tower = $22$ metres.

To Find:

The angle of elevation of the top of the tower from the eye of the observer.

Let the height of the tower be represented by AB and the position of the observer's eye by E. Let the point on the ground directly below the observer's eye be D. Let C be the foot of the tower.

The height of the tower AB = $22$ m.

The height of the observer ED = $1.5$ m.

The distance of the observer from the tower DC = $20.5$ m.

Draw a horizontal line EF from the observer's eye E parallel to the ground DC, meeting the tower AB at point F. Then EF = DC = $20.5$ m and FB = ED = $1.5$ m.

The height of the tower above the observer's eye level is AF = AB - FB.

$AF = 22 \text{ m} - 1.5 \text{ m} = 20.5 \text{ m}$.

The angle of elevation of the top of the tower from the eye of the observer is the angle $\angle$AEF. This is the angle $\theta$ in the right-angled triangle $\triangle$AEF, which is right-angled at F.

In $\triangle$AEF, we have:

Opposite side to $\theta$ (AF) = $20.5$ m

Adjacent side to $\theta$ (EF) = $20.5$ m

We use the tangent trigonometric ratio:

$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$

$\tan \theta = \frac{AF}{EF}$

Substitute the values:

$\tan \theta = \frac{20.5}{20.5}$

$\tan \theta = \cancel{\frac{20.5}{20.5}}^1$

$\tan \theta = 1$

We know that for an acute angle, $\tan \theta = 1$ implies $\theta = 45^\circ$.

$\tan \theta = \tan 45^\circ$

$\theta = 45^\circ$

The angle of elevation of the top of the tower from the eye of the observer is $\mathbf{45^\circ}$.

Solution:

We need to show that $\tan^4 \theta + \tan^2 \theta = \sec^4 \theta – \sec^2 \theta$.

Consider the Left Hand Side (LHS):

LHS $= \tan^4 \theta + \tan^2 \theta$

Factor out the common term $\tan^2 \theta$ from the LHS:

LHS $= \tan^2 \theta (\tan^2 \theta + 1)$

We use the fundamental trigonometric identity relating tangent and secant:

$\tan^2 \theta + 1 = \sec^2 \theta$

Substitute $\sec^2 \theta$ for $(\tan^2 \theta + 1)$ in the LHS expression:

LHS $= \tan^2 \theta (\sec^2 \theta)$

From the same identity, we can express $\tan^2 \theta$ in terms of $\sec^2 \theta$:

$\tan^2 \theta = \sec^2 \theta - 1$

Substitute this expression for $\tan^2 \theta$ into the current LHS expression:

LHS $= (\sec^2 \theta - 1) (\sec^2 \theta)$

Expand the expression by multiplying $\sec^2 \theta$ with each term inside the parenthesis:

LHS $= \sec^2 \theta \times \sec^2 \theta - 1 \times \sec^2 \theta$

LHS $= \sec^4 \theta - \sec^2 \theta$

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity $\tan^4 \theta + \tan^2 \theta = \sec^4 \theta – \sec^2 \theta$ is shown to be true for all values of $\theta$ where the functions are defined.

Let O be the eye of the observer, and C be the centre of the spherical balloon.

Let OX be the horizontal line through O. The angle of elevation of the centre C from O is $\angle COX = \phi$.

Let OA and OB be the tangents from O to the balloon at points A and B on its surface. The angle subtended by the balloon at O is $\angle AOB = \theta$.

Since C is the centre of the sphere and OA and OB are tangents from O, OC bisects $\angle AOB$. Also, the radius CA is perpendicular to the tangent OA at the point of contact A.

Therefore, in the right-angled triangle $\triangle OAC$:

$\angle AOC = \frac{\theta}{2}$

$\angle OAC = 90^\circ$

In $\triangle OAC$, we use the sine ratio:

$\sin(\angle AOC) = \frac{CA}{OC}$

Given the radius of the balloon is r, so $CA = r$.

$\sin\left(\frac{\theta}{2}\right) = \frac{r}{OC}$

Solving for OC (the distance from the observer's eye to the centre of the balloon):

Now, consider the vertical height of the centre C above the horizontal line OX. Let this height be $h$. Let P be the point directly below C on the line OX.

In the right-angled triangle $\triangle OPC$, we have:

$\angle OPC = 90^\circ$

$\angle COP = \phi$ (Angle of elevation)

CP = h (Height of the centre)

Using the sine ratio in $\triangle OPC$:

$\sin(\angle COP) = \frac{CP}{OC}$

$\sin(\phi) = \frac{h}{OC}$

Solving for h:

Substitute the expression for OC from equation (i) into equation (ii):

h = $\left(\frac{r}{\sin\left(\frac{\theta}{2}\right)}\right) \cdot \sin(\phi)$

h = $\frac{r \sin(\phi)}{\sin\left(\frac{\theta}{2}\right)}$

Therefore, the height of the centre of the balloon above the eye of the observer is $\mathbf{\frac{r \sin(\phi)}{\sin\left(\frac{\theta}{2}\right)}}$.

Let B be the position of the balloon and P be the point directly below the balloon on the straight road. Let $h$ be the height of the balloon, so $BP = h$.

Let A and C be the positions of the two cars on the road.

Let BX be the horizontal line through B.

The angles of depression from B to the cars A and C are given as $60^\circ$ and $45^\circ$. Let $\angle XBA$ and $\angle XBC$ be these angles.

By the property of alternate interior angles, the angles of elevation from the cars to the balloon are equal to the angles of depression.

So, the angles of elevation are $\angle BAP = 60^\circ$ and $\angle BCP = 45^\circ$ (or vice versa). Let's assume $\angle BAP = 60^\circ$ and $\angle BCP = 45^\circ$ without loss of generality. This implies car A is closer to P than car C.

In the right-angled triangle $\triangle BAP$:

$\tan(\angle BAP) = \frac{BP}{AP}$

The distance of car A from P is:

In the right-angled triangle $\triangle BCP$:

$\tan(\angle BCP) = \frac{BP}{CP}$

The distance of car C from P is:

We are given that the distance between the two cars is 100 m, i.e., $AC = 100$ m.

There are two possible cases for the positions of the cars on the straight road relative to P:

Case 1: The cars are on opposite sides of point P.

In this case, the distance between the cars is the sum of their distances from P.

Solving for h:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}-1$:

So, in this case, the height of the balloon is $\mathbf{50(3-\sqrt{3})}$ m.

Case 2: The cars are on the same side of point P.

From equations (i) and (ii), $AP = \frac{h}{\sqrt{3}}$ and $CP = h$. Since $\sqrt{3} > 1$, $AP < CP$. This means car A is closer to P than car C.

In this case, the distance between the cars is the difference of their distances from P.

Solving for h:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}+1$:

So, in this case, the height of the balloon is $\mathbf{50(3+\sqrt{3})}$ m.

Since the problem does not specify the relative positions of the cars, both $50(3-\sqrt{3})$ m and $50(3+\sqrt{3})$ m are possible heights depending on whether the cars are on opposite sides or the same side of the point directly below the balloon.

Given:

A point O is at a height $h$ metres above the surface of a lake.

Let L be the surface of the lake.

Let P be the point on the lake surface directly below O, so $OP = h$.

Let C be the position of the cloud above the lake surface.

Let M be the point on the lake surface directly below C.

The height of the cloud above the lake surface is $CM = H$ (say).

The reflection of the cloud C in the lake is C'. The reflection C' is as far below the lake surface as the cloud C is above it, so $MC' = CM = H$.

The angle of elevation of the cloud from O is $\theta$. Let OQ be a horizontal line from O meeting CM at Q. Then $\angle COQ = \theta$.

The angle of depression of the reflection C' from O is $\phi$. The angle between the horizontal line OQ and OC' is $\angle QOC' = \phi$.

To Prove:

The height of the cloud above the lake is $H = h \left( \frac{\tan\;\phi \;+\; \tan\;\theta}{\tan\;\phi\;-\; \tan\;\theta} \right)$.

Solution:

From the figure (imagined), we have OP = h. Since OQ is horizontal and PM is vertical, OQMP forms a rectangle. Thus, $QM = OP = h$ and $OQ = PM$.

Let $OQ = x$. Then $PM = x$.

In the right-angled triangle $\triangle COQ$:

The height $CQ = CM - QM = H - h$.

The base $OQ = x$.

From this, we can express x:

In the right-angled triangle $\triangle C'OQ$:

The height $C'Q = C'M + MQ = H + h$.

The base $OQ = x$.

From this, we can express x:

Equating the expressions for x from equation (i) and equation (ii):

Cross-multiply:

Expand both sides:

Rearrange the terms to group H terms on one side and h terms on the other:

Factor out H from the left side and h from the right side:

Finally, solve for H:

Thus, the height of the cloud above the lake is $h \left( \frac{\tan\;\phi \;+\; \tan\;\theta}{\tan\;\phi\;-\; \tan\;\theta} \right)$.

Hence Proved.

Given:

To Prove:

$\cos\; \theta = \frac{p^2 \;-\; 1}{p^2 \;+\; 1}$

Proof:

We know the identity involving $\text{cosec}\; \theta$ and $\cot\; \theta$ is:

Using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$), we can factor the left side:

Substitute the given value from equation (i):

From this, we get:

Now we have a system of two linear equations in terms of $\text{cosec}\; \theta$ and $\cot\; \theta$:

Equation (i): $\text{cosec}\; \theta + \cot\; \theta = p$

Equation (ii): $\text{cosec}\; \theta - \cot\; \theta = \frac{1}{p}$

Adding equation (i) and equation (ii):

Subtracting equation (ii) from equation (i):

We know that $\cos\; \theta = \frac{\cos\; \theta}{\sin\; \theta} / \frac{1}{\sin\; \theta} = \frac{\cot\; \theta}{\text{cosec}\; \theta}$.

Substitute the expressions for $\cot\; \theta$ and $\text{cosec}\; \theta$ from equations (iv) and (iii) respectively:

Cancel out the $2p$ term from the numerator and denominator:

Hence Proved.

To Prove:

$\sqrt{\sec^2\theta + \text{cosec}^2 \;\theta}$ = $\tan \theta + \cot \theta$

Proof:

Consider the Left Hand Side (LHS):

LHS = $\sqrt{\sec^2\theta + \text{cosec}^2 \;\theta}$

Using the trigonometric identities $\sec^2\theta = 1 + \tan^2\theta$ and $\text{cosec}^2\theta = 1 + \cot^2\theta$, we substitute them into the expression:

We know another fundamental identity: $\tan\theta \cdot \cot\theta = 1$.

So, we can write $2$ as $2 \cdot 1 = 2 (\tan\theta \cot\theta)$.

Substitute this into the expression under the square root:

This expression under the square root is in the form of $(a+b)^2 = a^2 + b^2 + 2ab$, where $a = \tan\theta$ and $b = \cot\theta$.

Taking the square root of a squared term:

For values of $\theta$ where $\tan\theta$ and $\cot\theta$ have the same sign (i.e., in quadrants I and III), $\tan\theta + \cot\theta$ is positive, so $|\tan\theta + \cot\theta| = \tan\theta + \cot\theta$.

Assuming the domain of $\theta$ is such that $\tan\theta + \cot\theta \ge 0$, we have:

This is the Right Hand Side (RHS).

Hence Proved.

Given:

Let AB be the tower of height h.

Let C be the initial position of the observer on the ground, and D be the new position after moving 20 metres towards the tower.

Let A be the base of the tower and B be the top.

The distance moved is $CD = 20$ m.

The initial angle of elevation from C to the top B is $\angle BCA = 30^\circ$.

The angle of elevation from D to the top B is increased by $15^\circ$, so $\angle BDA = 30^\circ + 15^\circ = 45^\circ$.

To Find:

The height of the tower, $h = AB$.

Solution:

Consider the right-angled triangle $\triangle ABD$:

Let the distance from the new observation point D to the base of the tower A be $AD = x$ metres.

In $\triangle ABD$, using the tangent function:

So,

Now, consider the right-angled triangle $\triangle ABC$:

The distance from the initial observation point C to the base A is $AC = AD + DC = x + 20$ metres.

In $\triangle ABC$, using the tangent function:

Substitute the value of $x$ from equation (i) into the equation from $\triangle ABC$:

Cross-multiply:

Rearrange the terms to isolate $h$:

Solve for $h$:

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3}+1$:

Therefore, the height of the tower is $\mathbf{10(\sqrt{3}+1)}$ metres.

Given:

To Prove:

$\tan \theta = 1$ or $\tan \theta = \frac{1}{2}$.

Proof:

Consider the given equation:

Assuming $\cos \theta \neq 0$, divide the entire equation by $\cos^2 \theta$ to convert it into terms of $\tan \theta$ and $\sec^2 \theta$:

Simplify each term using trigonometric identities ($\sec^2 \theta = \frac{1}{\cos^2 \theta}$, $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$, and $\tan \theta = \frac{\sin \theta}{\cos \theta}$):

Now, use the identity $\sec^2 \theta = 1 + \tan^2 \theta$:

Combine like terms:

Rearrange the equation into a quadratic equation in terms of $\tan \theta$:

Let $t = \tan \theta$. The quadratic equation becomes:

Factor the quadratic equation. We look for two numbers that multiply to $(2)(1) = 2$ and add up to $-3$. These numbers are $-2$ and $-1$.

Group terms and factor:

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1:

Since $t = \tan \theta$, we have $\tan \theta = \frac{1}{2}$.

Case 2:

Since $t = \tan \theta$, we have $\tan \theta = 1$.

Thus, the possible values for $\tan \theta$ are 1 and $\frac{1}{2}$.

Note: If $\cos \theta = 0$, then $\sin^2 \theta = 1$. The original equation becomes $1 + 1 = 3 \sin \theta \cdot 0$, which gives $2 = 0$, a contradiction. Hence, $\cos \theta$ cannot be 0, and dividing by $\cos^2 \theta$ is valid.

Hence Proved.

Given:

To Prove:

$2\sin \theta - \cos \theta = 2$

Proof:

Consider the given equation:

Square both sides of equation (i):

Expand the left side:

Use the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

Combine like terms:

Subtract 1 from both sides:

Factor out $\cos \theta$:

This equation implies either $\cos \theta = 0$ or $4\sin \theta + 3\cos \theta = 0$.

Case 1: $\cos \theta = 0$

Substitute $\cos \theta = 0$ into the given equation (i):

Now evaluate the expression $2\sin \theta - \cos \theta$ using $\sin \theta = 1$ and $\cos \theta = 0$:

In this case, $2\sin \theta - \cos \theta = 2$, which matches the required result.

Case 2: $4\sin \theta + 3\cos \theta = 0$

This implies $4\sin \theta = -3\cos \theta$. Assuming $\cos \theta \neq 0$, we can divide by $\cos \theta$:

If $\tan \theta = -3/4$, we can form a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is $\sqrt{3^2+4^2} = 5$. Since $\tan \theta$ is negative, $\theta$ lies in Quadrant II or Quadrant IV.

From the given equation $\sin \theta + 2\cos \theta = 1$, we can determine the signs of $\sin \theta$ and $\cos \theta$.

If $\theta$ is in Quadrant II: $\sin \theta > 0$, $\cos \theta < 0$. Based on $\tan \theta = -3/4$, we would have $\sin \theta = 3/5$ and $\cos \theta = -4/5$.

Check equation (i): $\sin \theta + 2\cos \theta = (3/5) + 2(-4/5) = 3/5 - 8/5 = -5/5 = -1$. This is not equal to 1. So Quadrant II is not possible.

If $\theta$ is in Quadrant IV: $\sin \theta < 0$, $\cos \theta > 0$. Based on $\tan \theta = -3/4$, we would have $\sin \theta = -3/5$ and $\cos \theta = 4/5$.

Check equation (i): $\sin \theta + 2\cos \theta = (-3/5) + 2(4/5) = -3/5 + 8/5 = 5/5 = 1$. This matches the given equation.

Now evaluate the expression $2\sin \theta - \cos \theta$ using $\sin \theta = -3/5$ and $\cos \theta = 4/5$ (the values that satisfy the given condition in this case):

In this case, $2\sin \theta - \cos \theta = -2$.

From the analysis of the given equation $\sin \theta + 2\cos \theta = 1$, we found that it is satisfied when $(\sin \theta = 1, \cos \theta = 0)$ or when $(\sin \theta = -3/5, \cos \theta = 4/5)$.

For the first set of values, $2\sin \theta - \cos \theta = 2$.

For the second set of values, $2\sin \theta - \cos \theta = -2$.

The given condition implies that $2\sin \theta - \cos \theta$ can be either 2 or -2.

However, the question asks to prove specifically that $2\sin \theta - \cos \theta = 2$. This result occurs when $\cos \theta = 0$ and $\sin \theta = 1$.

Hence Proved (that 2 is one of the possible values for $2\sin \theta - \cos \theta$ given the condition).

Given:

Let AB be the tower of height $h$. Let A be the foot of the tower and B be the top.

Let C and D be two points on the ground, in a straight line from the foot of the tower A.

The distance of point C from the foot is $AC = s$.

The distance of point D from the foot is $AD = t$.

The angle of elevation of the top B from C is $\angle BCA = \theta$.

The angle of elevation of the top B from D is $\angle BDA = \phi$.

The angles are complementary, which means $\theta + \phi = 90^\circ$.

To Prove:

The height of the tower is $h = \sqrt{st}$.

Proof:

Consider the right-angled triangle $\triangle ABC$:

In $\triangle ABC$, using the tangent function:

Now, consider the right-angled triangle $\triangle ABD$:

In $\triangle ABD$, using the tangent function:

We are given that the angles $\theta$ and $\phi$ are complementary. Thus, $\theta + \phi = 90^\circ$.

From this relationship, we can express $\phi$ as $\phi = 90^\circ - \theta$.

Substitute this into equation (ii):

Using the trigonometric identity $\tan(90^\circ - \theta) = \cot(\theta)$:

Now we have two expressions involving $\theta$ and $h$: one for $\tan(\theta)$ from equation (i) and one for $\cot(\theta)$ from equation (iii).

We know the reciprocal identity: $\tan(\theta) \cdot \cot(\theta) = 1$.

Substitute the expressions for $\tan(\theta)$ and $\cot(\theta)$ from equations (i) and (iii):

Multiply the terms on the left side:

Multiply both sides by $st$:

Take the square root of both sides. Since height ($h$), distance ($s$, $t$) are positive quantities, we take the positive square root:

Thus, the height of the tower is $\mathbf{\sqrt{st}}$.

Hence Proved.

Given:

Let AB be the tower of height $h$ metres standing on a level plane, where A is the foot and B is the top.

Let C and D be two points on the ground along the line of the shadow from the foot of the tower.

The angle of elevation of the top of the tower from point C is $\angle BCA = 60^\circ$.

The angle of elevation of the top of the tower from point D is $\angle BDA = 30^\circ$.

Since the angle of elevation is smaller at D ($30^\circ$) compared to C ($60^\circ$), point D is farther from the base of the tower A than point C.

The difference between the lengths of the shadows at these two points is given as 50 m. The shadow is longer at the lower elevation angle ($30^\circ$).

So, the length of the shadow at $30^\circ$ elevation (AD) is 50 m more than the length of the shadow at $60^\circ$ elevation (AC).

$AD - AC = 50$ m.

To Find:

The height of the tower, $h = AB$.

Solution:

Let the length of the shadow when the Sun's elevation is $60^\circ$ be $AC = x$ metres.

Then, the length of the shadow when the Sun's elevation is $30^\circ$ is $AD = x + 50$ metres.

Consider the right-angled triangle $\triangle ABC$, which is right-angled at A.

Using the tangent function in $\triangle ABC$:

From this equation, we can express $x$ in terms of $h$:

Now, consider the right-angled triangle $\triangle ABD$, which is right-angled at A.

Using the tangent function in $\triangle ABD$:

Substitute the expression for $x$ from equation (i) into equation (ii):

Simplify the denominator on the right side:

Rewrite the right side by multiplying by the reciprocal of the denominator:

Cross-multiply:

Rearrange the terms to solve for $h$. Subtract $h$ from both sides:

Divide by 2:

Therefore, the height of the tower is $\mathbf{25\sqrt{3}}$ metres.

Given:

Let AB be the vertical tower with A at the base and B at the top. Let the height of the tower be $H$.

A vertical flag staff BC of height $h$ is mounted on top of the tower, so C is the top of the flag staff.

Let P be a point on the horizontal plane from where the angles of elevation are observed. Let the distance from P to the base of the tower A be $x$.

The angle of elevation of the bottom of the flag staff (point B) from P is $\angle APB = \alpha$.

The angle of elevation of the top of the flag staff (point C) from P is $\angle APC = \beta$.

To Prove:

The height of the tower is $H = \left( \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \right)$.

Proof:

Consider the right-angled triangle $\triangle ABP$, which is right-angled at A.

In $\triangle ABP$, using the tangent function:

From this, we can express $x$ in terms of $H$ and $\tan \alpha$:

Now, consider the right-angled triangle $\triangle ACP$, which is right-angled at A.

The total height AC is the sum of the height of the tower and the height of the flag staff: $AC = AB + BC = H + h$.

In $\triangle ACP$, using the tangent function:

From this, we can express $x$ in terms of $H$, $h$, and $\tan \beta$:

Equate the expressions for $x$ from equation (i) and equation (ii):

Cross-multiply:

Distribute $\tan \alpha$ on the right side:

Collect terms involving H on the left side:

Factor out H from the left side:

Solve for H by dividing both sides by ($\tan \beta$ - $\tan \alpha$). We assume $\tan \beta \neq \tan \alpha$, which is true since $\beta > \alpha$ (as C is above B) and the angles are likely in the range $(0, 90^\circ)$.

Thus, the height of the tower is $\mathbf{\left( \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \right)}$.

Hence Proved.

Given:

To Prove:

$\sec \theta = \frac{l^2 \;+\; 1}{2l}$

Proof:

We know the fundamental trigonometric identity relating $\sec \theta$ and $\tan \theta$:

Using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$), we can factor the left side:

Substitute the given value from equation (i):

From this, we get:

Now we have a system of two linear equations in terms of $\sec \theta$ and $\tan \theta$:

Equation (i): $\sec \theta + \tan \theta = l$

Equation (ii): $\sec \theta - \tan \theta = \frac{1}{l}$

Add equation (i) and equation (ii):

Combine the terms on the right side by finding a common denominator:

Solve for $\sec \theta$ by dividing both sides by 2:

Thus, $\sec \theta = \mathbf{\frac{l^2 \;+\; 1}{2l}}$.

Hence Proved.

Given:

To Prove:

$q (p^2 – 1) = 2p$

Proof:

Consider the first given equation (i):

Square both sides of this equation:

Expand the left side using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

Use the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

Rearrange the equation to find an expression for $p^2 - 1$:

Now consider the second given equation (ii):

Rewrite $\sec \theta$ and $\text{cosec} \theta$ in terms of $\sin \theta$ and $\cos \theta$ using the reciprocal identities ($\sec \theta = \frac{1}{\cos \theta}$ and $\text{cosec} \theta = \frac{1}{\sin \theta}$):

Combine the fractions on the left side by finding a common denominator:

Now, consider the Left Hand Side (LHS) of the expression we need to prove: $q (p^2 – 1)$.

Substitute the expression for $q$ from equation (iv) and the expression for $(p^2 - 1)$ from equation (iii) into the LHS:

Notice that the term $\sin \theta \cos \theta$ appears in the denominator of the first factor and as a multiplier in the second factor. Assuming $\sin \theta \cos \theta \neq 0$, we can cancel this term:

Rearrange the terms:

From the given equation (i), we know that $\sin \theta + \cos \theta = p$. Substitute this into the expression:

This is the Right Hand Side (RHS) of the expression we needed to prove.

Thus, $q (p^2 – 1) = 2p$.

Note: The proof assumes $\sin \theta \neq 0$ and $\cos \theta \neq 0$, which means $\theta$ is not an integral multiple of $\frac{\pi}{2}$. If $\sin \theta = 0$ or $\cos \theta = 0$, then $\sec \theta$ or $\text{cosec} \theta$ would be undefined, and the variable $q$ would not exist. The problem statement implies that $q$ exists, hence $\sin \theta \neq 0$ and $\cos \theta \neq 0$.

Hence Proved.

Given:

To Prove:

a $\cos \theta$ – b $\sin \theta$ = $\sqrt{a^{2} + b^{2} - c^{2}}$

Proof:

Let $x = a \cos \theta – b \sin \theta$.

We want to show that $x = \sqrt{a^{2} + b^{2} - c^{2}}$. This is equivalent to showing that $x^2 = a^{2} + b^{2} - c^{2}$, provided $x \ge 0$. Or, more generally, that $(a \cos \theta – b \sin \theta)^2 = a^{2} + b^{2} - c^{2}$.

Square both sides of the given equation (i):

Expand the left side:

Now, consider the expression we want to find the value of, and square it:

Expand the left side:

Add equation (ii) and equation (iii):

Group the terms involving a$^2$ and b$^2$:

Factor out a$^2$ and b$^2$ and cancel the middle terms:

Use the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

Solve for x$^2$:

Take the square root of both sides:

Since $x = a \cos \theta – b \sin \theta$, we have shown that:

The problem asks to prove that a $\cos \theta$ – b $\sin \theta$ = $\sqrt{a^{2} + b^{2} - c^{2}}$. This indicates that we are interested in the positive value of the square root, or the context implies this result.

Hence Proved (that the square of the expression is $a^2 + b^2 - c^2$, implying the expression equals $\pm \sqrt{a^2 + b^2 - c^2}$).

To Prove:

$\frac{1 \;+\; \sec\; \theta \;-\; \tan\; \theta}{1 \;+\; \sec\;θ \;+\; \tan\; θ} = \frac{1 \;-\; \sin\; \theta}{\cos\; \theta}$

Proof:

Consider the Left Hand Side (LHS):

We know the identity $\sec^2 \theta - \tan^2 \theta = 1$. We can substitute $1$ in the numerator with this identity:

Factor the difference of squares term ($\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$):

Factor out the common term $(\sec \theta - \tan \theta)$ from the numerator:

Rearrange the terms inside the square brackets in the numerator:

Assuming that $1 + \sec \theta + \tan \theta \neq 0$, we can cancel the common factor in the numerator and denominator:

Now, convert $\sec \theta$ and $\tan \theta$ into terms of $\sin \theta$ and $\cos \theta$ using the identities $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

Since the fractions have a common denominator, combine them:

This is the Right Hand Side (RHS).

Hence Proved.

Given:

Let AB be the first tower with height $h_1 = 30$ m. A is the foot and B is the top.

Let CD be the second tower with height $h_2$. D is the foot and C is the top.

The two towers are on the same horizontal plane, and A and D are points on this plane. Let the distance between the feet of the towers be $d = AD$.

The angle of elevation of the top of the first tower (B) from the foot of the second tower (D) is $\angle BDA = 60^\circ$.

The angle of elevation of the top of the second tower (C) from the foot of the first tower (A) is $\angle CAD = 30^\circ$.

To Find:

The distance between the two towers, $d = AD$.

The height of the other tower, $h_2 = CD$.

Solution:

Consider the right-angled triangle $\triangle ABD$, which is right-angled at A.

The height of the tower AB is $h_1 = 30$ m.

The base AD is the distance between the towers, $d$.

Using the tangent function in $\triangle ABD$:

Solve for $d$:

Rationalize the denominator:

So, the distance between the two towers is $\mathbf{10\sqrt{3}}$ metres.

Now consider the right-angled triangle $\triangle CAD$, which is right-angled at D.

The height of the tower CD is $h_2$.

The base AD is the distance between the towers, $d = 10\sqrt{3}$ m.

Using the tangent function in $\triangle CAD$:

Solve for $h_2$:

Substitute the value of $d$ from equation (i) into equation (ii):

Cancel the $\sqrt{3}$ terms:

Therefore, the height of the other tower is $\mathbf{10}$ metres.

Given:

Let AB be the tower of height $h$ metres, where A is the foot and B is the top.

Let the two objects be at points C and D on the horizontal plane, in a line with the foot of the tower A.

Let BX be the horizontal line through B.

The angle of depression of object C from B is $\angle XBC = \alpha$.

The angle of depression of object D from B is $\angle XBD = \beta$.

We are given that $\beta > \alpha$. Since the angle of depression is larger for point D, D is closer to the foot of the tower A than point C.

By the property of alternate interior angles, the angles of elevation from C and D to B are equal to the angles of depression:

$\angle BCA = \angle XBC = \alpha$

$\angle BDA = \angle XBD = \beta$

To Find:

The distance between the two objects, $CD$.

Solution:

Let the distance from the foot of the tower to object D be $AD = x$ metres.

Let the distance from the foot of the tower to object C be $AC = y$ metres.

The distance between the two objects is $CD = AC - AD = y - x$.

Consider the right-angled triangle $\triangle ABD$, which is right-angled at A.

In $\triangle ABD$, using the tangent function:

Solve for $x$:

Now, consider the right-angled triangle $\triangle ABC$, which is right-angled at A.

In $\triangle ABC$, using the tangent function:

Solve for $y$:

The distance between the two objects is $CD = y - x$.

Substitute the expressions for $y$ from (ii) and $x$ from (i):

Factor out $h$:

Combine the terms inside the parentheses by finding a common denominator:

Therefore, the distance between the two objects is $\mathbf{h \left( \frac{\tan \beta - \tan \alpha}{\tan \alpha \tan \beta} \right)}$ metres.

Alternatively, using $\cot \theta = \frac{1}{\tan \theta}$:

This is an equivalent expression for the distance between the two objects: $\mathbf{h (\cot \alpha - \cot \beta)}$ metres.

Given:

A ladder of fixed length, let's say $L$, is leaning against a vertical wall.

Initial position: Angle with the horizontal is $\alpha$. Let the distance of the foot from the wall be $x_1$ and the height reached on the wall be $y_1$.

Final position: The foot is pulled away by distance $p$, so the new distance from the wall is $x_2 = x_1 + p$. The upper end slides down by distance $q$, so the new height reached on the wall is $y_2 = y_1 - q$. The new angle with the horizontal is $\beta$.

To Show:

$\frac{p}{q}$ = $\frac{\cos \;β \;-\; \cos\; α}{\sin\; α \;-\; \sin\; β}$

Proof:

Let $L$ be the length of the ladder. The ladder, the wall, and the ground form a right-angled triangle in both initial and final positions.

In the initial position, considering the right triangle formed by the ladder, the wall, and the ground:

From these equations, we can express the initial distances in terms of $L$ and $\alpha$:

In the final position, considering the new right triangle:

From these equations, we can express the final distances in terms of $L$ and $\beta$:

We are given that the foot is pulled away by a distance $p$, so $p = x_2 - x_1$. Substitute the expressions for $x_1$ and $x_2$:

We are also given that the upper end slides down by a distance $q$, so $q = y_1 - y_2$. Substitute the expressions for $y_1$ and $y_2$:

Now, form the ratio $\frac{p}{q}$ using equations (i) and (ii):

Assuming $L \neq 0$ (the ladder exists) and $\sin \alpha - \sin \beta \neq 0$ (the height changes), we can cancel $L$ from the numerator and the denominator:

Thus, $\mathbf{\frac{p}{q} = \frac{cos \;β \;-\; cos\; α}{sin\; α \;-\; sin\; β}}$.

Hence Proved.

Given:

Let AB be the vertical tower with height $h$. Let A be the foot of the tower on the ground and B be the top.

Let C be the first point on the ground from where the angle of elevation is observed. Let $AC = x$ be the distance from the foot of the tower to point C.

The angle of elevation of the top of the tower from C is $\angle BCA = 60^\circ$.

Let D be the second point, located 10 m vertically above C. So $CD = 10$ m.

Draw a horizontal line from D meeting AB at E. Since D is vertically above C and AE is part of the vertical tower, ACDE forms a rectangle. Thus, $AE = CD = 10$ m and $DE = AC = x$.

The height of the tower above point D is $BE = AB - AE = h - 10$.

The angle of elevation of the top of the tower (B) from D is $\angle BDE = 45^\circ$.

To Find:

The height of the tower, $h = AB$.

Solution:

Consider the right-angled triangle $\triangle ABC$, which is right-angled at A.

Using the tangent function in $\triangle ABC$:

Solving for x:

Now, consider the right-angled triangle $\triangle BDE$, which is right-angled at E.

In $\triangle BDE$, using the tangent function:

Equate the expressions for x from equation (i) and equation (ii):

Multiply both sides by $\sqrt{3}$:

Rearrange the terms to solve for h:

Solve for h:

To rationalize the denominator, multiply the numerator and denominator by the conjugate ($\sqrt{3}+1$):

Therefore, the height of the tower is $\mathbf{5(3+\sqrt{3})}$ metres.

Given:

Let W be the position of the window, and let A be the point on the ground directly below the window. The height of the window from the ground is $WA = h$ metres.

Let CD be the other house, where D is the bottom (foot) and C is the top. The house CD is situated on the opposite side of a lane from the first house.

Let AD be the width of the lane.

Draw a horizontal line WE from the window W, parallel to the ground AD, meeting the second house at point E. Then WE is perpendicular to CD, and WE = AD.

The angle of elevation of the top of the other house (C) from the window (W) is $\angle EWC = \alpha$.

The angle of depression of the bottom of the other house (D) from the window (W) is $\angle EWD = \beta$.

Since WE is parallel to AD, and WD is a transversal, the alternate interior angle $\angle WDA = \angle EWD = \beta$.

Also, since WEAD is a rectangle, $ED = WA = h$.

The height of the other house is $CD = CE + ED$. We know $ED = h$, so $CD = CE + h$.

To Prove:

The height of the other house is $CD = h ( 1 + \tan \alpha \cot \beta )$ metres.

Proof:

Consider the right-angled triangle $\triangle WED$, which is right-angled at E.

In $\triangle WED$, we have:

From this, we can find the width of the lane, $WE$ (or $AD$). Assuming $\tan \beta \neq 0$:

Using the identity $\frac{1}{\tan \beta} = \cot \beta$:

Now, consider the right-angled triangle $\triangle WEC$, which is right-angled at E.

In $\triangle WEC$, we have:

From this, we can find the height CE:

Substitute the expression for $WE$ from equation (i) into equation (ii):

The total height of the other house is $CD = CE + ED$.

Substitute the expression for CE from equation (iii) and $ED = h$:

Factor out $h$ from the right side:

Rearrange the terms inside the parentheses:

Thus, the height of the other house is $\mathbf{h ( 1 + \tan α \cot β )}$ metres.

Hence Proved.

Given:

Let the ground level be the horizontal reference plane.

Let L be the position of the lower window, and U be the position of the upper window.

The height of the lower window from the ground is 2 m.

The upper window is 4 m vertically above the lower window, so the height of the upper window from the ground is $2 + 4 = 6$ m.

Let B be the position of the balloon above the ground. Let P be the point on the ground directly below the balloon. Let the height of the balloon above the ground be $H$ metres, so $BP = H$.

Draw a horizontal line from the lower window L, meeting BP at Q. The height of Q from the ground is equal to the height of the lower window, i.e., $QP = 2$ m.

Draw a horizontal line from the upper window U, meeting BP at R. The height of R from the ground is equal to the height of the upper window, i.e., $RP = 6$ m.

The vertical distance from L to the balloon is $BQ = BP - QP = H - 2$.

The vertical distance from U to the balloon is $BR = BP - RP = H - 6$.

The horizontal distance from the windows to the vertical line below the balloon is the same. Let this distance be $x$. So, $LQ = UR = x$.

The angle of elevation of the balloon from the lower window (L) is $\angle BLQ = 60^\circ$.

The angle of elevation of the balloon from the upper window (U) is $\angle BUR = 30^\circ$.

To Find:

The height of the balloon above the ground, $H$.

Solution:

Consider the right-angled triangle $\triangle BLQ$, which is right-angled at Q.

In $\triangle BLQ$, using the tangent function:

Solving for x:

Now, consider the right-angled triangle $\triangle BUR$, which is right-angled at R.

In $\triangle BUR$, using the tangent function:

Solving for x:

Equate the expressions for x from equation (i) and equation (ii):

Multiply both sides by $\sqrt{3}$:

Rearrange the terms to solve for H. Subtract H from both sides and add 18 to both sides:

Divide by 2:

Therefore, the height of the balloon above the ground is $\mathbf{8}$ metres.