Chapter 8 Introduction To Trigonometry

Given

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

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To Find

The other trigonometric ratios of the angle A.

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Solution

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

In a right triangle, the trigonometric ratio $\tan A$ is defined as the ratio of the side opposite to angle A to the side adjacent to angle A.

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

We are given $\tan A = \frac{4}{3}$.

So, $\frac{BC}{AB} = \frac{4}{3}$.

Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

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

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

$AC^2 = 25k^2$

$AC = \sqrt{25k^2} = 5k$

(Since length is positive)

Now we can find the other trigonometric ratios:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$

$\text{cosec } A = \frac{1}{\sin A} = \frac{\text{Hypotenuse}}{\text{Side opposite to angle A}} = \frac{AC}{BC} = \frac{5k}{4k} = \frac{5}{4}$

$\sec A = \frac{1}{\cos A} = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}} = \frac{AC}{AB} = \frac{5k}{3k} = \frac{5}{3}$

$\cot A = \frac{1}{\tan A} = \frac{\text{Side adjacent to angle A}}{\text{Side opposite to angle A}} = \frac{AB}{BC} = \frac{3k}{4k} = \frac{3}{4}$

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The other trigonometric ratios of angle A are:

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

$\cos A = \frac{3}{5}$

$\text{cosec } A = \frac{5}{4}$

$\sec A = \frac{5}{3}$

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

Given

$\angle$B and $\angle$Q are acute angles.

$\sin B = \sin Q$.

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To Prove

$\angle B = \angle Q$.

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Proof

Consider two right-angled triangles, $\triangle ABC$ and $\triangle PQR$, where $\angle C = 90^\circ$ and $\angle R = 90^\circ$.

By the definition of the sine ratio:

In $\triangle ABC$, $\sin B = \frac{\text{Side opposite to }\angle B}{\text{Hypotenuse}} = \frac{AC}{AB}$

In $\triangle PQR$, $\sin Q = \frac{\text{Side opposite to }\angle Q}{\text{Hypotenuse}} = \frac{PR}{PQ}$

We are given that $\sin B = \sin Q$.

Let $\frac{AC}{AB} = \frac{PR}{PQ} = k$, where $k$ is a positive constant.

Then, $AC = k \cdot AB$ and $PR = k \cdot PQ$.

Using the Pythagorean theorem in $\triangle ABC$:

$BC^2 = AB^2 - AC^2$

$BC = \sqrt{AB^2 - AC^2} = \sqrt{AB^2 - (k \cdot AB)^2} = \sqrt{AB^2(1 - k^2)} \ $$ = AB \sqrt{1 - k^2}$

Using the Pythagorean theorem in $\triangle PQR$:

$QR^2 = PQ^2 - PR^2$

$QR = \sqrt{PQ^2 - PR^2} = \sqrt{PQ^2 - (k \cdot PQ)^2} = \sqrt{PQ^2(1 - k^2)} \ $$ = PQ \sqrt{1 - k^2}$

Now consider the ratio of the adjacent sides, BC and QR:

$\frac{BC}{QR} = \frac{AB \sqrt{1 - k^2}}{PQ \sqrt{1 - k^2}} = \frac{AB}{PQ}$

From equation (i), by rearranging the terms, we get $\frac{AC}{PR} = \frac{AB}{PQ}$.

So, we have the ratios of corresponding sides equal:

Since the ratios of all three pairs of corresponding sides are equal, $\triangle ABC$ is similar to $\triangle PQR$ by the SSS similarity criterion.

When two triangles are similar, their corresponding angles are equal.

Therefore, $\angle B = \angle Q$.

Hence Proved.

Given:

In $\triangle$ ACB, $\angle$C = $90^\circ$, AB = 29 units, BC = 21 units, and $\angle$ABC = $\theta$.

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To Find:

(i) $\cos^2 \theta + \sin^2 \theta$

(ii) $\cos^2 \theta - \sin^2 \theta$

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Solution:

In the right-angled $\triangle$ ACB, by the Pythagorean theorem, we have:

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

$(29)^2 = AC^2 + (21)^2$

$841 = AC^2 + 441$

$AC^2 = 841 - 441$

$AC^2 = 400$

$AC = \sqrt{400} = 20$ units (Since length is positive)

Now, we determine the trigonometric ratios $\sin \theta$ and $\cos \theta$ with respect to angle $\theta$ ($\angle$ABC):

$\sin \theta = \frac{\text{Side opposite to }\theta}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{20}{29}$

$\cos \theta = \frac{\text{Side adjacent to }\theta}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{21}{29}$

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(i) Value of $\cos^2 \theta + \sin^2 \theta$:

$\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2$

$= \frac{21^2}{29^2} + \frac{20^2}{29^2}$

$= \frac{441}{841} + \frac{400}{841}$

$= \frac{441 + 400}{841}$

$= \frac{841}{841} = 1$

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

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(ii) Value of $\cos^2 \theta - \sin^2 \theta$:

$\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2$

$= \frac{21^2}{29^2} - \frac{20^2}{29^2}$

$= \frac{441}{841} - \frac{400}{841}$

$= \frac{441 - 400}{841}$

$= \frac{41}{841}$

So, $\cos^2 \theta - \sin^2 \theta = \frac{41}{841}$.

Given:

In $\triangle$ ABC, $\angle$B = $90^\circ$ and $\tan A = 1$.

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To Verify:

$2 \sin A \cos A = 1$.

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Solution:

In a right triangle ABC, right-angled at B, the trigonometric ratio $\tan A$ is defined as the ratio of the side opposite to angle A to the side adjacent to angle A.

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

We are given $\tan A = 1$.

So, $\frac{BC}{AB} = 1$, which implies $BC = AB$.

Let $AB = BC = k$, where $k$ is a positive real number.

By the Pythagorean theorem in $\triangle$ ABC, the hypotenuse AC is:

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

$AC^2 = k^2 + k^2$

$AC^2 = 2k^2$

$AC = \sqrt{2k^2} = k\sqrt{2}$ (Since length is positive)

Now, we find the values of $\sin A$ and $\cos A$:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$

Now, we evaluate the expression $2 \sin A \cos A$:

$2 \sin A \cos A = 2 \times \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{\sqrt{2}}\right)$

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

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

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

Thus, we have verified that $2 \sin A \cos A = 1$ when $\tan A = 1$ in a right-angled triangle at B.

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Alternate Solution:

We are given $\tan A = 1$.

We know that $\tan 45^\circ = 1$. Since A is an acute angle in a right triangle, this implies that $A = 45^\circ$.

Now, we find $\sin A$ and $\cos A$ for $A = 45^\circ$:

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

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

Substitute these values into the expression $2 \sin A \cos A$:

$2 \sin A \cos A = 2 \times \sin 45^\circ \times \cos 45^\circ$

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

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

$= 1$

Thus, $2 \sin A \cos A = 1$ is verified.

Given:

In $\triangle$ OPQ, $\angle$P = $90^\circ$, OP = 7 cm and OQ – PQ = 1 cm.

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To Find:

The values of $\sin$ Q and $\cos$ Q.

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Solution:

In the right-angled $\triangle$ OPQ, we are given:

From equation (i), we can express OQ in terms of PQ:

Applying the Pythagorean theorem to $\triangle$ OPQ:

$OQ^2 = OP^2 + PQ^2$

Substitute the values from the given information and the relation between OQ and PQ:

$(PQ + 1)^2 = (7)^2 + PQ^2$

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

$PQ^2 + 2(PQ)(1) + 1^2 = 49 + PQ^2$

$PQ^2 + 2PQ + 1 = 49 + PQ^2$

Subtract $PQ^2$ from both sides:

$2PQ + 1 = 49$

Subtract 1 from both sides:

$2PQ = 49 - 1$

$2PQ = 48$

Divide by 2:

$PQ = \frac{48}{2}$

$PQ = 24$ cm

Now substitute the value of PQ into the relation OQ = PQ + 1:

$OQ = 24 + 1$

$OQ = 25$ cm

So, the lengths of the sides are OP = 7 cm (Opposite to angle Q), PQ = 24 cm (Adjacent to angle Q), and OQ = 25 cm (Hypotenuse).

Now, we can find the values of $\sin$ Q and $\cos$ Q:

$\sin Q = \frac{\text{Side opposite to angle Q}}{\text{Hypotenuse}} = \frac{OP}{OQ}$

$\sin Q = \frac{7}{25}$

$\cos Q = \frac{\text{Side adjacent to angle Q}}{\text{Hypotenuse}} = \frac{PQ}{OQ}$

$\cos Q = \frac{24}{25}$

The values are $\sin Q = \frac{7}{25}$ and $\cos Q = \frac{24}{25}$.

Given

A right-angled triangle ∆ ABC, with the right angle at B.

Length of side AB = 24 cm.

Length of side BC = 7 cm.

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To Find

(i) The values of sin A and cos A.

(ii) The values of sin C and cos C.

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Solution

First, we need to find the length of the hypotenuse AC. Since ∆ ABC is a right-angled triangle, we can use the Pythagorean theorem.

According to the Pythagorean theorem:

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

Substituting the given values:

$AC^2 = (24)^2 + (7)^2$

$AC^2 = 576 + 49$

$AC^2 = 625$

$AC = \sqrt{625}$

$AC = 25$ cm

So, the length of the hypotenuse AC is 25 cm.

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(i) To determine sin A and cos A

For angle A:

The side opposite to angle A is BC = 7 cm.

The side adjacent to angle A is AB = 24 cm.

The hypotenuse is AC = 25 cm.

$\sin A = \frac{\text{Side opposite to A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

$\cos A = \frac{\text{Side adjacent to A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

Thus, $\sin A = \frac{7}{25}$ and $\cos A = \frac{24}{25}$.

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(ii) To determine sin C and cos C

For angle C:

The side opposite to angle C is AB = 24 cm.

The side adjacent to angle C is BC = 7 cm.

The hypotenuse is AC = 25 cm.

$\sin C = \frac{\text{Side opposite to C}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

$\cos C = \frac{\text{Side adjacent to C}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

Thus, $\sin C = \frac{24}{25}$ and $\cos C = \frac{7}{25}$.

Given:

In $\triangle$ PQR, $\angle$Q = $90^\circ$, PQ = 12 cm, PR = 13 cm.

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To Find:

$\tan$ P – $\cot$ R.

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Solution:

In the right-angled $\triangle$ PQR, by the Pythagorean theorem, we have:

$PR^2 = PQ^2 + QR^2$

$(13)^2 = (12)^2 + QR^2$

$169 = 144 + QR^2$

$QR^2 = 169 - 144$

$QR^2 = 25$

$QR = \sqrt{25}$

$QR = 5$ cm (Since length must be positive)

Now, we find the values of $\tan$ P and $\cot$ R.

For angle P:

The side opposite to angle P is QR.

The side adjacent to angle P is PQ.

$\tan P = \frac{\text{Side opposite to angle P}}{\text{Side adjacent to angle P}} = \frac{QR}{PQ} = \frac{5}{12}$

For angle R:

The side opposite to angle R is PQ.

The side adjacent to angle R is QR.

$\cot R = \frac{\text{Side adjacent to angle R}}{\text{Side opposite to angle R}} = \frac{QR}{PQ} = \frac{5}{12}$

Now, we calculate $\tan$ P – $\cot$ R:

$\tan P - \cot R = \frac{5}{12} - \frac{5}{12}$

$\tan P - \cot R = 0$

Given

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

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To Find

The values of $\cos A$ and $\tan A$.

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Solution

We are given $\sin A = \frac{3}{4}$.

In a right-angled triangle, $\sin A$ is defined as the ratio of the side opposite to angle A to the hypotenuse.

Let's consider a right triangle ABC, right-angled at B.

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC}$

So, $\frac{BC}{AC} = \frac{3}{4}$.

Let $BC = 3k$ and $AC = 4k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

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

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

$AB^2 = 16k^2 - 9k^2$

$AB^2 = 7k^2$

$AB = \sqrt{7k^2} = k\sqrt{7}$

Now, we can find $\cos A$ and $\tan A$:

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\cos A = \frac{k\sqrt{7}}{4k} = \frac{\sqrt{7}}{4}$

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

$\tan A = \frac{3k}{k\sqrt{7}} = \frac{3}{\sqrt{7}}$

Thus, $\cos A = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{3}{\sqrt{7}}$.

Given

$15 \cot A = 8$

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To Find

The values of $\sin A$ and $\sec A$.

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Solution

We are given the equation $15 \cot A = 8$.

Dividing both sides by 15, we get:

$\cot A = \frac{8}{15}$

In a right-angled triangle, the trigonometric ratio $\cot A$ is defined as the ratio of the side adjacent to angle A to the side opposite to angle A.

Let's consider a right triangle ABC, right-angled at B.

$\cot A = \frac{\text{Side adjacent to angle A}}{\text{Side opposite to angle A}} = \frac{AB}{BC}$

So, $\frac{AB}{BC} = \frac{8}{15}$.

Let $AB = 8k$ and $BC = 15k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

$AC^2 = (8k)^2 + (15k)^2$

$AC^2 = 64k^2 + 225k^2$

$AC^2 = 289k^2$

$AC = \sqrt{289k^2} = 17k$

Now, we can find $\sin A$ and $\sec A$:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\sin A = \frac{15k}{17k} = \frac{15}{17}$

$\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}} = \frac{AC}{AB}$

$\sec A = \frac{17k}{8k} = \frac{17}{8}$

Thus, $\sin A = \frac{15}{17}$ and $\sec A = \frac{17}{8}$.

Given

$\sec \theta = \frac{13}{12}$

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To Find

All other trigonometric ratios for angle $\theta$.

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Solution

We are given $\sec \theta = \frac{13}{12}$.

In a right-angled triangle, $\sec \theta$ is defined as the ratio of the hypotenuse to the side adjacent to angle $\theta$.

Let's consider a right triangle ABC, right-angled at B, with $\angle A = \theta$.

$\sec \theta = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle }\theta} = \frac{AC}{AB}$

So, $\frac{AC}{AB} = \frac{13}{12}$.

Let $AC = 13k$ and $AB = 12k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

$(13k)^2 = (12k)^2 + BC^2$

$169k^2 = 144k^2 + BC^2$

$BC^2 = 169k^2 - 144k^2$

$BC^2 = 25k^2$

$BC = \sqrt{25k^2} = 5k$

Now, we can find the other trigonometric ratios:

$\sin \theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$

$\cos \theta = \frac{\text{Side adjacent to angle }\theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$

$\tan \theta = \frac{\text{Side opposite to angle }\theta}{\text{Side adjacent to angle }\theta} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$

$\text{cosec } \theta = \frac{\text{Hypotenuse}}{\text{Side opposite to angle }\theta} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$

$\cot \theta = \frac{\text{Side adjacent to angle }\theta}{\text{Side opposite to angle }\theta} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$

The other trigonometric ratios are:

$\sin \theta = \frac{5}{13}$, $\cos \theta = \frac{12}{13}$, $\tan \theta = \frac{5}{12}$, $\text{cosec } \theta = \frac{13}{5}$, $\cot \theta = \frac{12}{5}$

Given

$\angle$A and $\angle$B are acute angles.

$\cos A = \cos B$.

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To Show

$\angle A = \angle B$.

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Proof

Consider a right-angled triangle ABC, where $\angle C = 90^\circ$. In this triangle, A and B are the acute angles.

By the definition of the cosine ratio:

For angle A, $\cos A = \frac{\text{Side adjacent to }\angle A}{\text{Hypotenuse}} = \frac{AC}{AB}$

For angle B, $\cos B = \frac{\text{Side adjacent to }\angle B}{\text{Hypotenuse}} = \frac{BC}{AB}$

We are given that $\cos A = \cos B$.

Multiplying both sides by AB, we get:

$AC = BC$

In $\triangle$ ABC, we know that angles opposite to equal sides are equal.

The angle opposite to side AC is $\angle B$.

The angle opposite to side BC is $\angle A$.

Since AC = BC, their opposite angles must be equal.

Hence Shown.

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Alternate Proof

Let's consider two different right-angled triangles, $\triangle APQ$ and $\triangle BMN$, where $\angle P = 90^\circ$ and $\angle M = 90^\circ$.

Given $\cos A = \cos B$.

$\cos A = \frac{AP}{AQ}$ and $\cos B = \frac{BM}{BN}$

So, $\frac{AP}{AQ} = \frac{BM}{BN} = k$ (let)

This implies $AP = k \cdot AQ$ and $BM = k \cdot BN$.

By Pythagorean theorem:

$PQ = \sqrt{AQ^2 - AP^2} = \sqrt{AQ^2 - (k \cdot AQ)^2} = AQ\sqrt{1-k^2}$

$MN = \sqrt{BN^2 - BM^2} = \sqrt{BN^2 - (k \cdot BN)^2} = BN\sqrt{1-k^2}$

Now, consider the ratio $\frac{PQ}{MN}$:

$\frac{PQ}{MN} = \frac{AQ\sqrt{1-k^2}}{BN\sqrt{1-k^2}} = \frac{AQ}{BN}$

From our initial assumption, $\frac{AP}{AQ} = \frac{BM}{BN}$, which can be rearranged to $\frac{AP}{BM} = \frac{AQ}{BN}$.

Thus we have $\frac{AP}{BM} = \frac{AQ}{BN} = \frac{PQ}{MN}$.

By SSS similarity criterion, $\triangle APQ \sim \triangle BMN$.

Since the triangles are similar, their corresponding angles are equal.

Therefore, $\angle A = \angle B$.

Hence Shown.

Given

$\cot \theta = \frac{7}{8}$

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To Evaluate

(i) $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$

(ii) $\cot^2 \theta$

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Solution

(i) Evaluate $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$:

We simplify the expression using the algebraic identity $(a+b)(a-b) = a^2 - b^2$.

Numerator: $(1 + \sin \theta)(1 - \sin \theta) = 1^2 - \sin^2 \theta = 1 - \sin^2 \theta$

Denominator: $(1 + \cos \theta)(1 - \cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta$

The expression becomes: $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we know that $1 - \sin^2 \theta = \cos^2 \theta$ and $1 - \cos^2 \theta = \sin^2 \theta$.

Substituting these into the expression:

$\frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = (\cot \theta)^2 = \cot^2 \theta$

We are given $\cot \theta = \frac{7}{8}$.

So, the value of the expression is $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

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(ii) Evaluate $\cot^2 \theta$:

We are given $\cot \theta = \frac{7}{8}$.

$\cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$

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The final answers are:

(i) $\frac{49}{64}$

(ii) $\frac{49}{64}$

Given

$3 \cot A = 4$

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To Check

Whether the identity $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ is true for the given value.

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Solution

From the given equation, $3 \cot A = 4$, we get $\cot A = \frac{4}{3}$.

Since $\tan A = \frac{1}{\cot A}$, we have $\tan A = \frac{3}{4}$.

Let's construct a right-angled triangle ABC, right-angled at B.

$\cot A = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{AB}{BC} = \frac{4}{3}$.

Let $AB = 4k$ and $BC = 3k$.

By the Pythagorean theorem: $AC^2 = AB^2 + BC^2 \ $$ = (4k)^2 + (3k)^2 \ $$ = 16k^2 + 9k^2 = 25k^2$.

So, $AC = \sqrt{25k^2} = 5k$.

Now, we find $\sin A$ and $\cos A$:

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

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

Now, we evaluate the Left Hand Side (LHS) of the identity:

LHS = $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{(16-9)/16}{(16+9)/16} = \frac{7/16}{25/16} = \frac{7}{25}$

Next, we evaluate the Right Hand Side (RHS):

RHS = $\cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25}$

Since LHS = RHS = $\frac{7}{25}$, the given statement is true.

Yes, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$.

Given

In $\triangle$ ABC, $\angle B = 90^\circ$ and $\tan A = \frac{1}{\sqrt{3}}$.

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To Find

(i) The value of $\sin A \cos C + \cos A \sin C$.

(ii) The value of $\cos A \cos C – \sin A \sin C$.

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Solution

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

In $\triangle$ ABC, $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{1}{\sqrt{3}}$.

Let $BC = k$ and $AB = k\sqrt{3}$.

By Pythagorean theorem: $AC^2 = AB^2 + BC^2 \ $$ = (k\sqrt{3})^2 + k^2 \ $$ = 3k^2 + k^2 = 4k^2$.

So, $AC = \sqrt{4k^2} = 2k$.

Now, we find the required trigonometric ratios for angles A and C:

$\sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

$\cos A = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}$

$\sin C = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}$

$\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

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(i) Evaluate $\sin A \cos C + \cos A \sin C$:

Substituting the values:

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

$= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$

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(ii) Evaluate $\cos A \cos C – \sin A \sin C$:

Substituting the values:

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

$= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$

---

The final values are:

(i) 1

(ii) 0

Given

In $\triangle PQR$, $\angle Q = 90^\circ$.

PQ = 5 cm.

PR + QR = 25 cm.

---

To Find

The values of $\sin P$, $\cos P$, and $\tan P$.

---

Solution

In the right-angled $\triangle PQR$, by the Pythagorean theorem:

We are given PR + QR = 25 cm, so we can write PR = 25 – QR.

Substitute PR = 25 – QR and PQ = 5 into equation (i):

$(25 – QR)^2 = 5^2 + QR^2$

$625 - 50QR + QR^2 = 25 + QR^2$

Subtracting $QR^2$ from both sides:

$625 – 50QR = 25$

$625 – 25 = 50QR$

$600 = 50QR$

$QR = \frac{600}{50} = 12$ cm

Now, find PR:

PR = 25 – QR = 25 – 12 = 13 cm

So, we have the sides: PQ = 5 cm (Adjacent to P), QR = 12 cm (Opposite to P), and PR = 13 cm (Hypotenuse).

Now, we can determine the required trigonometric ratios for angle P:

$\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$

$\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$

$\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{12}{5}$

---

The required values are:

$\sin P = \frac{12}{13}$, $\cos P = \frac{5}{13}$, and $\tan P = \frac{12}{5}$.

(i) The value of tan A is always less than 1.

False.

Justification:

In a right-angled triangle, $\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}}$.

The length of the side opposite to angle A can be greater than, equal to, or less than the length of the side adjacent to angle A, depending on the value of angle A.

For example, if $A = 45^\circ$, the opposite side equals the adjacent side, so $\tan 45^\circ = 1$.

If $A > 45^\circ$ (and $A < 90^\circ$), the side opposite to A is greater than the side adjacent to A, so $\tan A > 1$. For instance, $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is greater than 1.

Therefore, the value of $\tan A$ is not always less than 1.

---

(ii) sec A = $\frac{12}{5}$ for some value of angle A.

True.

Justification:

In a right-angled triangle, $\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}}$.

The hypotenuse is always the longest side in a right-angled triangle (except for the degenerate case of a right angle with zero area, where sides can be zero, but for an angle in a triangle, sides must be positive). Thus, the ratio of the hypotenuse to the adjacent side must be greater than 1.

Given $\sec A = \frac{12}{5} = 2.4$. Since $2.4 > 1$, this value is a valid value for $\sec A$ for some acute angle A.

We can form a right triangle with adjacent side 5 units and hypotenuse 12 units. By the Pythagorean theorem, the opposite side would be $\sqrt{12^2 - 5^2} = \sqrt{144 - 25} = \sqrt{119}$ units. Since $\sqrt{119}$ is a real positive number, such a triangle exists, and $\sec A = \frac{12}{5}$ is possible.

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(iii) cos A is the abbreviation used for the cosecant of angle A.

False.

Justification:

$\cos A$ is the abbreviation used for the cosine of angle A.

The abbreviation used for the cosecant of angle A is $\text{cosec } A$ (or $\text{csc } A$).

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(iv) cot A is the product of cot and A.

False.

Justification:

'cot' is part of the trigonometric ratio name 'cotangent'. $\cot A$ represents the cotangent of the angle A. It is not a product of 'cot' and the angle A. The expression $\cot A$ is a single entity representing the ratio of the adjacent side to the opposite side for the angle A in a right triangle.

---

(v) sin θ = $\frac{4}{3}$ for some angle θ.

False.

Justification:

In a right-angled triangle, $\sin \theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}}$.

The length of the side opposite to any angle in a right triangle is always less than or equal to the length of the hypotenuse. Therefore, the ratio $\frac{\text{Opposite}}{\text{Hypotenuse}}$ must be less than or equal to 1.

Given $\sin \theta = \frac{4}{3} \approx 1.33$. Since $1.33 > 1$, this value is not possible for $\sin \theta$ for any real angle $\theta$. The range of the sine function is $[-1, 1]$. For angles in a triangle (which are between $0^\circ$ and $180^\circ$), the value of sine is between 0 and 1 (inclusive).

Given

In a right-angled triangle ∆ABC:

The right angle is at B, so $\angle B = 90^\circ$.

The length of the side AB = 5 cm.

The measure of angle ACB, $\angle C = 30^\circ$.

---

To Find

The lengths of the sides BC and AC.

---

Solution

We can use trigonometric ratios to find the lengths of the unknown sides.

With respect to the given angle $\angle C = 30^\circ$:

Side AB is the Opposite Side.

Side BC is the Adjacent Side.

Side AC is the Hypotenuse.

---

Finding the length of BC:

We need to find the adjacent side (BC) and we know the opposite side (AB). The trigonometric ratio that relates the opposite and adjacent sides is the tangent (tan).

$\tan C = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{AB}{BC}$

Substituting the given values:

$\tan 30^\circ = \frac{5}{BC}$

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

$\frac{1}{\sqrt{3}} = \frac{5}{BC}$

By cross-multiplication, we get:

$BC = 5\sqrt{3}$ cm.

---

Finding the length of AC:

We need to find the hypotenuse (AC) and we know the opposite side (AB). The trigonometric ratio that relates the opposite side and the hypotenuse is the sine (sin).

$\sin C = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

Substituting the given values:

$\sin 30^\circ = \frac{5}{AC}$

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

$\frac{1}{2} = \frac{5}{AC}$

By cross-multiplication, we get:

$AC = 5 \times 2 = 10$ cm.

---

Therefore, the lengths of the sides are BC = $5\sqrt{3}$ cm and AC = 10 cm.

Given:

In $\triangle$ PQR, $\angle$Q = $90^\circ$, PQ = 3 cm, and PR = 6 cm.

---

To Determine:

$\angle$QPR and $\angle$PRQ.

---

Solution:

In the right-angled $\triangle$ PQR, we have PQ = 3 cm and PR = 6 cm. The hypotenuse is PR.

Consider angle P ($\angle$QPR):

The side adjacent to angle P is PQ.

The hypotenuse is PR.

We can use the cosine ratio:

$\cos P = \frac{\text{Side adjacent to P}}{\text{Hypotenuse}} = \frac{PQ}{PR}$

$\cos P = \frac{3}{6} = \frac{1}{2}$

We know that $\cos 60^\circ = \frac{1}{2}$.

Since P is an acute angle in a right triangle, comparing the values, we get:

$P = 60^\circ$

So, $\angle$QPR = $60^\circ$.

Consider angle R ($\angle$PRQ):

The side opposite to angle R is PQ.

The hypotenuse is PR.

We can use the sine ratio:

$\sin R = \frac{\text{Side opposite to R}}{\text{Hypotenuse}} = \frac{PQ}{PR}$

$\sin R = \frac{3}{6} = \frac{1}{2}$

We know that $\sin 30^\circ = \frac{1}{2}$.

Since R is an acute angle in a right triangle, comparing the values, we get:

$R = 30^\circ$

So, $\angle$PRQ = $30^\circ$.

Alternatively, once we found $\angle$P = $60^\circ$, we can find $\angle$R using the angle sum property of a triangle:

In $\triangle$ PQR, $\angle$P + $\angle$Q + $\angle$R = $180^\circ$

$60^\circ + 90^\circ + \angle$R = $180^\circ$

$150^\circ + \angle$R = $180^\circ$

$\angle$R = $180^\circ - 150^\circ = 30^\circ$.

So, $\angle$PRQ = $30^\circ$.

The angles are $\angle$QPR = $60^\circ$ and $\angle$PRQ = $30^\circ$.

Given:

$\sin (A - B) = \frac{1}{2}$

$\cos (A + B) = \frac{1}{2}$

$0^\circ < A + B \leq 90^\circ$

$A > B$

---

To Find:

The values of angles A and B.

---

Solution:

We are given the equation $\sin (A - B) = \frac{1}{2}$.

We know that $\sin 30^\circ = \frac{1}{2}$.

Since the sine function is one-to-one for angles in the range typically considered in these problems (especially given A > B and the context of acute angles), we can equate the arguments:

We are also given the equation $\cos (A + B) = \frac{1}{2}$.

We know that $\cos 60^\circ = \frac{1}{2}$.

The condition $0^\circ < A + B \leq 90^\circ$ tells us that A + B is an acute angle or a right angle in the first quadrant boundary. Since $\cos (A+B) = \frac{1}{2}$ (which is positive), A + B must be in the first quadrant.

Therefore, we can equate the arguments:

Now we have a system of two linear equations with two variables A and B:

Equation (i): $A - B = 30^\circ$

Equation (ii): $A + B = $60^\circ$

Add equation (i) and equation (ii):

$(A - B) + (A + B) = 30^\circ + 60^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

$A = 45^\circ$

Substitute the value of A into equation (ii):

$45^\circ + B = 60^\circ$

$B = 60^\circ - 45^\circ$

$B = 15^\circ$

Let's check if these values satisfy the given conditions:

$A = 45^\circ$, $B = 15^\circ$.

$A > B$: $45^\circ > 15^\circ$ (True)

$0^\circ < A + B \leq 90^\circ$: $A + B = 45^\circ + 15^\circ = 60^\circ$. $0^\circ < 60^\circ \leq 90^\circ$ (True)

The values of A and B are $45^\circ$ and $15^\circ$ respectively.

(i) Evaluate $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$:

We use the standard trigonometric values:

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

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

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

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

Substitute these values into the expression:

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

$= \frac{(\sqrt{3})^2}{2^2} + \frac{1^2}{2^2}$

$= \frac{3}{4} + \frac{1}{4}$

$= \frac{3 + 1}{4} = \frac{4}{4} = 1$

The value is 1.

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(ii) Evaluate $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$:

We use the standard trigonometric values:

$\tan 45^\circ = 1$

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

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

Substitute these values into the expression:

$2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ = 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$

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

$= 2 + 0 = 2$

The value is 2.

---

(iii) Evaluate $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$:

We use the standard trigonometric values:

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

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$

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

Substitute these values into the expression:

$\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$

Find a common denominator in the denominator:

$= \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}}$

Invert the denominator and multiply:

$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}}$

$= \frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})}$

$= \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{2} - 2\sqrt{6}$ or $2(\sqrt{2} - \sqrt{6})$. It's simpler to factor out 2 first.

$= \frac{\sqrt{3}}{2(\sqrt{2} + \sqrt{6})}$

Now multiply by $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$ (conjugate of $\sqrt{2} + \sqrt{6}$ after swapping terms for easier calculation of difference of squares):

$= \frac{\sqrt{3}(\sqrt{6}-\sqrt{2})}{2(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}$

$= \frac{\sqrt{18}-\sqrt{6}}{2((\sqrt{6})^2-(\sqrt{2})^2)}$

$= \frac{\sqrt{9 \times 2}-\sqrt{6}}{2(6-2)}$

$= \frac{3\sqrt{2}-\sqrt{6}}{2(4)}$

$= \frac{3\sqrt{2}-\sqrt{6}}{8}$

The value is $\frac{3\sqrt{2}-\sqrt{6}}{8}$.

---

(iv) Evaluate $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$:

We use the standard trigonometric values:

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

$\tan 45^\circ = 1$

$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$

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

$\cot 45^\circ = 1$

Substitute these values into the expression:

Numerator = $\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ = \frac{1}{2} + 1 - \frac{2}{\sqrt{3}}$

$= \frac{1}{2} + \frac{2}{2} - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}$

Find a common denominator: $\frac{3\sqrt{3} - 4}{2\sqrt{3}}$

Denominator = $\sec 30^\circ + \cos 60^\circ + \cot 45^\circ = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1$

$= \frac{2}{\sqrt{3}} + \frac{1}{2} + \frac{2}{2} = \frac{2}{\sqrt{3}} + \frac{3}{2}$

Find a common denominator: $\frac{4 + 3\sqrt{3}}{2\sqrt{3}}$

Now divide the numerator by the denominator:

$\frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}} \times \frac{2\sqrt{3}}{4 + 3\sqrt{3}}$

$= \frac{3\sqrt{3} - 4}{\cancel{2\sqrt{3}}} \times \frac{\cancel{2\sqrt{3}}}{3\sqrt{3} + 4}$

$= \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $3\sqrt{3} - 4$:

$= \frac{(3\sqrt{3} - 4)(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)}$

Use $(a-b)^2 = a^2 - 2ab + b^2$ for the numerator and $(a+b)(a-b) = a^2 - b^2$ for the denominator:

Numerator = $(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2 = (9 \times 3) - 24\sqrt{3} + 16 \ $$ = 27 - 24\sqrt{3} + 16 \ $$ = 43 - 24\sqrt{3}$

Denominator = $(3\sqrt{3})^2 - (4)^2 = (9 \times 3) - 16 = 27 - 16 = 11$

The expression is equal to $\frac{43 - 24\sqrt{3}}{11}$.

---

(v) Evaluate $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$:

We use the standard trigonometric values:

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

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

$\tan 45^\circ = 1$

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

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

Substitute these values into the expression:

Numerator = $5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ = 5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2$

$= 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1$

$= \frac{5}{4} + \frac{16}{3} - 1$

Find a common denominator (12):

$= \frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{12}$

$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12}$

$= \frac{15 + 64 - 12}{12} = \frac{79 - 12}{12} = \frac{67}{12}$

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

$= \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1$

Alternatively, using the identity $\sin^2 \theta + \cos^2 \theta = 1$, the denominator is simply 1.

Now divide the numerator by the denominator:

$\frac{\frac{67}{12}}{1} = \frac{67}{12}$

The value is $\frac{67}{12}$.

(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$:

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

Substitute this value into the expression:

$\frac{2 (\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$= \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

$= \frac{2}{\sqrt{3}} \times \frac{3}{4}$

$= \frac{\cancel{2}^1}{\sqrt{3}} \times \frac{3}{\cancel{4}^2} = \frac{3}{2\sqrt{3}}$

Rationalize the denominator:

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

Now, we compare this value with the given options:

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

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

(C) $\tan 60^\circ = \sqrt{3}$

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

The value of the expression is equal to $\sin 60^\circ$.

The correct option is (A).

Justification: Calculation shows the expression evaluates to $\frac{\sqrt{3}}{2}$, which is the value of $\sin 60^\circ$.

---

(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$:

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

Substitute this value into the expression:

$\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$

Now, we compare this value with the given options:

(A) $\tan 90^\circ$ (Undefined)

(B) 1

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

(D) 0

The value of the expression is 0.

The correct option is (D).

Justification: Calculation shows the expression evaluates to 0.

---

(iii) $\sin 2A = 2 \sin A$ is true when A =:

We need to check which value of A satisfies the equation $\sin 2A = 2 \sin A$.

(A) If $A = 0^\circ$:

LHS = $\sin (2 \times 0^\circ) = \sin 0^\circ = 0$

RHS = $2 \sin 0^\circ = 2 \times 0 = 0$

LHS = RHS. So, A = $0^\circ$ is a solution.

(B) If $A = 30^\circ$:

LHS = $\sin (2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$

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

LHS $\neq$ RHS.

(C) If $A = 45^\circ$:

LHS = $\sin (2 \times 45^\circ) = \sin 90^\circ = 1$

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

LHS $\neq$ RHS.

(D) If $A = 60^\circ$:

LHS = $\sin (2 \times 60^\circ) = \sin 120^\circ$. $\sin 120^\circ \ $$ = \sin(180^\circ - 60^\circ) \ $$ = \sin 60^\circ \ $$ = \frac{\sqrt{3}}{2}$

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

LHS $\neq$ RHS.

The equation $\sin 2A = 2 \sin A$ is true only when $A = 0^\circ$ (or other values not typically covered in basic trigonometry). Based on the options provided, only A = $0^\circ$ satisfies the condition.

The correct option is (A).

Justification: We checked each option and found that only A = $0^\circ$ satisfies the equation $\sin 2A = 2 \sin A$. (Using the double angle identity $\sin 2A = 2 \sin A \cos A$, the equation becomes $2 \sin A \cos A = 2 \sin A$, which simplifies to $2 \sin A \cos A - 2 \sin A = 0$, or $2 \sin A (\cos A - 1) = 0$. This is true if $\sin A = 0$ or $\cos A = 1$. Both occur when $A = 0^\circ$.)

---

(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$:

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

Substitute this value into the expression:

$\frac{2 (\frac{1}{\sqrt{3}})}{1 - (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}}$

$= \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$= \frac{2}{\sqrt{3}} \times \frac{3}{2}$

$= \frac{\cancel{2}}{\sqrt{3}} \times \frac{3}{\cancel{2}} = \frac{3}{\sqrt{3}}$

Rationalize the denominator:

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

Now, we compare this value with the given options:

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

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

(C) $\tan 60^\circ = \sqrt{3}$

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

The value of the expression is equal to $\tan 60^\circ$.

The correct option is (C).

Justification: Calculation shows the expression evaluates to $\sqrt{3}$, which is the value of $\tan 60^\circ$. (This expression is also the formula for $\tan 2A$, so $\tan(2 \times 30^\circ) = \tan 60^\circ$).

Given:

$\tan (A + B) = \sqrt{3}$

$\tan (A - B) = \frac{1}{\sqrt{3}}$

$0^\circ < A + B \leq 90^\circ$

$A > B$

---

To Find:

The values of angles A and B.

---

Solution:

We are given the equation $\tan (A + B) = \sqrt{3}$.

We know that $\tan 60^\circ = \sqrt{3}$.

Given that $0^\circ < A + B \leq 90^\circ$, and the tangent is positive, A + B must be in the first quadrant. The unique angle in this range whose tangent is $\sqrt{3}$ is $60^\circ$.

Therefore, we can equate the arguments:

We are also given the equation $\tan (A - B) = \frac{1}{\sqrt{3}}$.

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

Given that $A > B$, the difference $A - B$ must be positive. Also, since $A+B \leq 90^\circ$ and $B > 0$ (otherwise $A=A+B$ and $\tan A = \sqrt{3}$ would mean $A=60^\circ$, then $B=0$, and $\tan(60-0) = \tan 60 = \sqrt{3} \neq 1/\sqrt{3}$), it implies $A < 90^\circ$ and $B$ is also an acute angle, making $A-B$ also likely acute. The unique angle whose tangent is $\frac{1}{\sqrt{3}}$ is $30^\circ$.

Therefore, we can equate the arguments:

Now we have a system of two linear equations with two variables A and B:

From equation (i): $A + B = 60^\circ$

From equation (ii): $A - B = 30^\circ$

Add equation (i) and equation (ii):

$(A + B) + (A - B) = 60^\circ + 30^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

A = $45^\circ$

Substitute the value of A = $45^\circ$ into equation (i):

$45^\circ + B = 60^\circ$

$B = 60^\circ - 45^\circ$

B = $15^\circ$

Let's verify if these values satisfy the given conditions:

$A = 45^\circ$ and $B = 15^\circ$.

$A > B$: $45^\circ > 15^\circ$. This condition is satisfied.

$0^\circ < A + B \leq 90^\circ$: $A + B = 45^\circ + 15^\circ = 60^\circ$. $0^\circ < 60^\circ \leq 90^\circ$. This condition is satisfied.

The values of A and B are $45^\circ$ and $15^\circ$ respectively.

(i) $\sin (A + B) = \sin A + \sin B$.

False.

Justification:

Consider A = $30^\circ$ and B = $60^\circ$.

LHS = $\sin (A + B) = \sin (30^\circ + 60^\circ) = \sin 90^\circ = 1$.

RHS = $\sin A + \sin B = \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$.

Since $1 \neq \frac{1+\sqrt{3}}{2}$, the statement $\sin (A + B) = \sin A + \sin B$ is generally false.

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(ii) The value of $\sin \theta$ increases as $\theta$ increases.

True (for $0^\circ \leq \theta \leq 90^\circ$).

Justification:

Consider the values of $\sin \theta$ for increasing $\theta$ in the range from $0^\circ$ to $90^\circ$ (which is the typical range for angles in a right triangle as considered in this chapter):

$\sin 0^\circ = 0$

$\sin 30^\circ = 0.5$

$\sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

$\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

$\sin 90^\circ = 1$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\sin \theta$ monotonically increases from 0 to 1. Therefore, the statement is true for this range of $\theta$.

---

(iii) The value of $\cos \theta$ increases as $\theta$ increases.

False.

Justification:

Consider the values of $\cos \theta$ for increasing $\theta$ in the range from $0^\circ$ to $90^\circ$:

$\cos 0^\circ = 1$

$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

$\cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

$\cos 60^\circ = 0.5$

$\cos 90^\circ = 0$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\cos \theta$ monotonically decreases from 1 to 0. Therefore, the statement is false for this range of $\theta$.

---

(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.

False.

Justification:

The equality $\sin \theta = \cos \theta$ is only true for specific values of $\theta$, such as $\theta = 45^\circ$ (or $45^\circ + n \times 180^\circ$, where $n$ is an integer). It is not true for all values of $\theta$.

For example, if $\theta = 30^\circ$, $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Since $\frac{1}{2} \neq \frac{\sqrt{3}}{2}$, the statement is false.

---

(v) $\cot A$ is not defined for $A = 0^\circ$.

True.

Justification:

The trigonometric ratio $\cot A$ is defined as $\cot A = \frac{\cos A}{\sin A}$.

For $A = 0^\circ$, we have $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$.

So, $\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$.

Division by zero is undefined. Therefore, $\cot A$ is not defined for $A = 0^\circ$.

To Evaluate:

$\frac{\tan 65^\circ}{\cot 25^\circ}$

---

Solution:

We use the trigonometric identity for complementary angles: $\tan (90^\circ - \theta) = \cot \theta$ and $\cot (90^\circ - \theta) = \tan \theta$.

Notice that $65^\circ + 25^\circ = 90^\circ$. So, $65^\circ$ and $25^\circ$ are complementary angles.

We can write $65^\circ = 90^\circ - 25^\circ$.

So, $\tan 65^\circ = \tan (90^\circ - 25^\circ)$.

Using the identity $\tan (90^\circ - \theta) = \cot \theta$ with $\theta = 25^\circ$, we get:

$\tan 65^\circ = \cot 25^\circ$

Alternatively, we can write $25^\circ = 90^\circ - 65^\circ$.

So, $\cot 25^\circ = \cot (90^\circ - 65^\circ)$.

Using the identity $\cot (90^\circ - \theta) = \tan \theta$ with $\theta = 65^\circ$, we get:

$\cot 25^\circ = \tan 65^\circ$

Now, substitute either of these results back into the expression:

$\frac{\tan 65^\circ}{\cot 25^\circ} = \frac{\cot 25^\circ}{\cot 25^\circ} = 1$ (using $\tan 65^\circ = \cot 25^\circ$)

or

$\frac{\tan 65^\circ}{\cot 25^\circ} = \frac{\tan 65^\circ}{\tan 65^\circ} = 1$ (using $\cot 25^\circ = \tan 65^\circ$)

The value of the expression is 1.

Given:

$\sin 3A = \cos (A – 26^\circ)$

3A is an acute angle ($0^\circ < 3A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

We are given the equation $\sin 3A = \cos (A – 26^\circ)$.

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

Since 3A is an acute angle, we can write $\sin 3A$ as $\cos (90^\circ - 3A)$.

Substitute this into the given equation:

$\cos (90^\circ - 3A) = \cos (A – 26^\circ)$

Since both angles are likely acute (or within the range where cosine is one-to-one or its values correspond uniquely to acute angles in this context), we can equate the arguments:

$90^\circ - 3A = A – 26^\circ$

Rearrange the terms to solve for A:

$90^\circ + 26^\circ = A + 3A$

$116^\circ = 4A$

$A = \frac{116^\circ}{4}$

$A = 29^\circ$

Let's check if the conditions are satisfied with A = $29^\circ$:

3A = $3 \times 29^\circ = 87^\circ$. This is an acute angle ($0^\circ < 87^\circ < 90^\circ$), so the condition is satisfied.

A – $26^\circ = 29^\circ – 26^\circ = 3^\circ$. This angle is also acute and positive.

With $A = 29^\circ$, the original equation is $\sin 87^\circ = \cos 3^\circ$. This is true because $\sin 87^\circ = \sin (90^\circ - 3^\circ) = \cos 3^\circ$.

The value of A is $29^\circ$.

To Express:

$\cot 85^\circ + \cos 75^\circ$ in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

---

Solution:

We need to express the given trigonometric ratios of angles $85^\circ$ and $75^\circ$ (which are greater than $45^\circ$) in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

We use the complementary angle identities:

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

$\cos \theta = \sin (90^\circ - \theta)$

For the term $\cot 85^\circ$:

The angle is $85^\circ$. The complementary angle is $90^\circ - 85^\circ = 5^\circ$.

$\cot 85^\circ = \cot (90^\circ - 5^\circ) = \tan 5^\circ$

The angle $5^\circ$ is between $0^\circ$ and $45^\circ$.

For the term $\cos 75^\circ$:

The angle is $75^\circ$. The complementary angle is $90^\circ - 75^\circ = 15^\circ$.

$\cos 75^\circ = \cos (90^\circ - 15^\circ) = \sin 15^\circ$

The angle $15^\circ$ is between $0^\circ$ and $45^\circ$.

Now, substitute these expressions back into the original sum:

$\cot 85^\circ + \cos 75^\circ = \tan 5^\circ + \sin 15^\circ$

The expression $\cot 85^\circ + \cos 75^\circ$ is expressed in terms of trigonometric ratios of angles $5^\circ$ and $15^\circ$, both of which are between $0^\circ$ and $45^\circ$.

The expression is $\tan 5^\circ + \sin 15^\circ$.

(i) Evaluate $\frac{\sin 18^\circ}{\cos 72^\circ}$:

We notice that $18^\circ + 72^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ or $\cos (90^\circ - \theta) = \sin \theta$.

Let's express the numerator in terms of cosine:

$\sin 18^\circ = \sin (90^\circ - 72^\circ) = \cos 72^\circ$

Substitute this into the expression:

$\frac{\sin 18^\circ}{\cos 72^\circ} = \frac{\cos 72^\circ}{\cos 72^\circ}$

Assuming $\cos 72^\circ \neq 0$, which is true since $0^\circ < 72^\circ < 90^\circ$, we can cancel the terms.

$= 1$

Alternatively, express the denominator in terms of sine:

$\cos 72^\circ = \cos (90^\circ - 18^\circ) = \sin 18^\circ$

Substitute this into the expression:

$\frac{\sin 18^\circ}{\cos 72^\circ} = \frac{\sin 18^\circ}{\sin 18^\circ}$

Assuming $\sin 18^\circ \neq 0$, which is true since $0^\circ < 18^\circ < 90^\circ$, we can cancel the terms.

$= 1$

The value is 1.

---

(ii) Evaluate $\frac{\tan 26^\circ}{\cot 64^\circ}$:

We notice that $26^\circ + 64^\circ = 90^\circ$. These are complementary angles.

We use the identity $\tan (90^\circ - \theta) = \cot \theta$ or $\cot (90^\circ - \theta) = \tan \theta$.

Let's express the numerator in terms of cotangent:

$\tan 26^\circ = \tan (90^\circ - 64^\circ) = \cot 64^\circ$

Substitute this into the expression:

$\frac{\tan 26^\circ}{\cot 64^\circ} = \frac{\cot 64^\circ}{\cot 64^\circ}$

Assuming $\cot 64^\circ \neq 0$, which is true since $0^\circ < 64^\circ < 90^\circ$, we can cancel the terms.

$= 1$

The value is 1.

---

(iii) Evaluate $\cos 48^\circ – \sin 42^\circ$:

We notice that $48^\circ + 42^\circ = 90^\circ$. These are complementary angles.

We use the identity $\cos (90^\circ - \theta) = \sin \theta$ or $\sin (90^\circ - \theta) = \cos \theta$.

Let's express the first term in terms of sine:

$\cos 48^\circ = \cos (90^\circ - 42^\circ) = \sin 42^\circ$

Substitute this into the expression:

$\cos 48^\circ – \sin 42^\circ = \sin 42^\circ – \sin 42^\circ$

$= 0$

Alternatively, express the second term in terms of cosine:

$\sin 42^\circ = \sin (90^\circ - 48^\circ) = \cos 48^\circ$

Substitute this into the expression:

$\cos 48^\circ – \sin 42^\circ = \cos 48^\circ – \cos 48^\circ$

$= 0$

The value is 0.

---

(iv) Evaluate $\text{cosec } 31^\circ – \sec 59^\circ$:

We notice that $31^\circ + 59^\circ = 90^\circ$. These are complementary angles.

We use the identity $\text{cosec } (90^\circ - \theta) = \sec \theta$ or $\sec (90^\circ - \theta) = \text{cosec } \theta$.

Let's express the first term in terms of secant:

$\text{cosec } 31^\circ = \text{cosec } (90^\circ - 59^\circ) = \sec 59^\circ$

Substitute this into the expression:

$\text{cosec } 31^\circ – \sec 59^\circ = \sec 59^\circ – \sec 59^\circ$

$= 0$

Alternatively, express the second term in terms of cosecant:

$\sec 59^\circ = \sec (90^\circ - 31^\circ) = \text{cosec } 31^\circ$

Substitute this into the expression:

$\text{cosec } 31^\circ – \sec 59^\circ = \text{cosec } 31^\circ – \text{cosec } 31^\circ$

$= 0$

The value is 0.

(i) Show that $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$:

Consider the Left Hand Side (LHS):

LHS = $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$

Group the terms that are complementary:

Notice that $48^\circ + 42^\circ = 90^\circ$ and $23^\circ + 67^\circ = 90^\circ$.

We use the identity $\tan (90^\circ - \theta) = \cot \theta$.

$\tan 48^\circ = \tan (90^\circ - 42^\circ) = \cot 42^\circ$

$\tan 23^\circ = \tan (90^\circ - 67^\circ) = \cot 67^\circ$

Substitute these into the LHS:

LHS = $(\cot 42^\circ) (\cot 67^\circ) (\tan 42^\circ) (\tan 67^\circ)$

Rearrange the terms:

LHS = $(\cot 42^\circ \tan 42^\circ) (\cot 67^\circ \tan 67^\circ)$

We know that $\cot \theta \tan \theta = 1$ (since $\cot \theta = \frac{1}{\tan \theta}$).

LHS = $(1) \times (1)$

LHS = 1

This is equal to the Right Hand Side (RHS).

Hence Shown.

---

(ii) Show that $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ = 0$:

Consider the Left Hand Side (LHS):

LHS = $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ$

Notice that $38^\circ + 52^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ and $\cos (90^\circ - \theta) = \sin \theta$.

Let's express the terms involving $52^\circ$ in terms of $38^\circ$:

$\cos 52^\circ = \cos (90^\circ - 38^\circ) = \sin 38^\circ$

$\sin 52^\circ = \sin (90^\circ - 38^\circ) = \cos 38^\circ$

Substitute these into the LHS:

LHS = $\cos 38^\circ (\sin 38^\circ) – \sin 38^\circ (\cos 38^\circ)$

LHS = $\cos 38^\circ \sin 38^\circ – \sin 38^\circ \cos 38^\circ$

The two terms are identical but with opposite signs, so they cancel out.

LHS = 0

This is equal to the Right Hand Side (RHS).

Hence Shown.

---

Alternate Solution for (ii) (using angle sum identity):

Recall the cosine addition formula: $\cos (A + B) = \cos A \cos B - \sin A \sin B$.

The expression $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ$ matches the form of $\cos (A + B)$ with $A = 38^\circ$ and $B = 52^\circ$.

LHS = $\cos (38^\circ + 52^\circ)$

LHS = $\cos 90^\circ$

We know that $\cos 90^\circ = 0$.

LHS = 0

This is equal to the RHS.

Hence Shown.

Given:

$\tan 2A = \cot (A – 18^\circ)$

2A is an acute angle ($0^\circ < 2A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

We are given the equation $\tan 2A = \cot (A – 18^\circ)$.

We know the trigonometric identity for complementary angles: $\tan \theta = \cot (90^\circ - \theta)$.

Since 2A is an acute angle, we can write $\tan 2A$ as $\cot (90^\circ - 2A)$.

Substitute this into the given equation:

$\cot (90^\circ - 2A) = \cot (A – 18^\circ)$

Since $2A$ is an acute angle, $90^\circ - 2A$ is also an angle between $0^\circ$ and $90^\circ$. For these angles, the cotangent function is one-to-one. Thus, we can equate the arguments of the cotangent function:

$90^\circ - 2A = A – 18^\circ$

Now, we solve this linear equation for A.

Add $2A$ to both sides:

$90^\circ = A – 18^\circ + 2A$

$90^\circ = 3A – 18^\circ$

Add $18^\circ$ to both sides:

$90^\circ + 18^\circ = 3A$

$108^\circ = 3A$

Divide by 3:

$A = \frac{108^\circ}{3}$

$A = 36^\circ$

Let's verify the condition that 2A is an acute angle:

$2A = 2 \times 36^\circ = 72^\circ$. Since $0^\circ < 72^\circ < 90^\circ$, 2A is indeed an acute angle.

Also, $A - 18^\circ = 36^\circ - 18^\circ = 18^\circ$, which is positive.

The value of A is $36^\circ$.

Given:

$\tan A = \cot B$

Angles A and B are acute angles (implied by the context of trigonometric ratios in a right triangle in this section).

---

To Prove:

A + B = $90^\circ$.

---

Solution:

We are given the equation $\tan A = \cot B$.

We use the trigonometric identity for complementary angles: $\cot \theta = \tan (90^\circ - \theta)$.

We can rewrite the right side of the given equation using this identity:

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

Substitute this back into the given equation:

Since A and B are acute angles (and thus $90^\circ - B$ is also an angle between $0^\circ$ and $90^\circ$), and the tangent function is one-to-one for angles between $0^\circ$ and $90^\circ$, we can equate the arguments:

$A = 90^\circ - B$

Add B to both sides of the equation:

$A + B = 90^\circ$

Thus, if $\tan A = \cot B$ and A and B are acute angles, then A + B = $90^\circ$.

Hence Proved.

---

Alternate Solution:

We are given $\tan A = \cot B$.

We can use the identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot B = \frac{\cos B}{\sin B}$.

$\frac{\sin A}{\cos A} = \frac{\cos B}{\sin B}$

Cross-multiply:

$\sin A \sin B = \cos A \cos B$

Rearrange the terms:

$\cos A \cos B - \sin A \sin B = 0$

Recognize the left side as the cosine addition formula $\cos (A + B) = \cos A \cos B - \sin A \sin B$:

$\cos (A + B) = 0$

For acute angles A and B, A + B will be in the range $(0^\circ, 180^\circ)$. The angle in this range whose cosine is 0 is $90^\circ$.

Therefore, $A + B = 90^\circ$.

Hence Proved.

Given:

$\sec 4A = \text{cosec } (A – 20^\circ)$

4A is an acute angle ($0^\circ < 4A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

We are given the equation $\sec 4A = \text{cosec } (A – 20^\circ)$.

We use the trigonometric identity for complementary angles: $\sec \theta = \text{cosec } (90^\circ - \theta)$ or $\text{cosec } \theta = \sec (90^\circ - \theta)$.

Since 4A is an acute angle, we can write $\sec 4A$ as $\text{cosec } (90^\circ - 4A)$.

Substitute this into the given equation:

$\text{cosec } (90^\circ - 4A) = \text{cosec } (A – 20^\circ)$

Since $4A$ is an acute angle, $90^\circ - 4A$ is also an angle between $0^\circ$ and $90^\circ$. For these angles, the cosecant function is one-to-one. Thus, we can equate the arguments of the cosecant function:

$90^\circ - 4A = A – 20^\circ$

Now, we solve this linear equation for A.

Add $4A$ to both sides:

$90^\circ = A – 20^\circ + 4A$

$90^\circ = 5A – 20^\circ$

Add $20^\circ$ to both sides:

$90^\circ + 20^\circ = 5A$

$110^\circ = 5A$

Divide by 5:

$A = \frac{110^\circ}{5}$

$A = 22^\circ$

Let's verify the condition that 4A is an acute angle:

$4A = 4 \times 22^\circ = 88^\circ$. Since $0^\circ < 88^\circ < 90^\circ$, 4A is indeed an acute angle.

Also, $A - 20^\circ = 22^\circ - 20^\circ = 2^\circ$, which is positive.

The value of A is $22^\circ$.

Given:

A, B, and C are the interior angles of a triangle ABC.

---

To Show:

$\sin \left( \frac{B + C}{2} \right) = \cos \frac{A}{2}$

---

Solution:

Since A, B, and C are the interior angles of a triangle, their sum is $180^\circ$.

We need the expression $\frac{B+C}{2}$. Let's isolate B + C from the equation:

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

Now, divide both sides of the equation by 2:

$\frac{B + C}{2} = \frac{180^\circ - A}{2}$

$\frac{B + C}{2} = \frac{180^\circ}{2} - \frac{A}{2}$

$\frac{B + C}{2} = 90^\circ - \frac{A}{2}$

Now, take the sine of both sides of this equation:

$\sin \left( \frac{B + C}{2} \right) = \sin \left( 90^\circ - \frac{A}{2} \right)$

We use the trigonometric identity for complementary angles: $\sin (90^\circ - \theta) = \cos \theta$.

Here, $\theta = \frac{A}{2}$.

So, $\sin \left( 90^\circ - \frac{A}{2} \right) = \cos \frac{A}{2}$.

Substitute this back into the equation:

$\sin \left( \frac{B + C}{2} \right) = \cos \frac{A}{2}$

This matches the required identity.

Hence Shown.

To Express:

$\sin 67^\circ + \cos 75^\circ$ in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

---

Solution:

We need to express the given trigonometric ratios of angles $67^\circ$ and $75^\circ$ (which are greater than $45^\circ$) in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

We use the complementary angle identities:

$\sin \theta = \cos (90^\circ - \theta)$

$\cos \theta = \sin (90^\circ - \theta)$

For the term $\sin 67^\circ$:

The angle is $67^\circ$. The complementary angle is $90^\circ - 67^\circ = 23^\circ$.

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

The angle $23^\circ$ is between $0^\circ$ and $45^\circ$.

For the term $\cos 75^\circ$:

The angle is $75^\circ$. The complementary angle is $90^\circ - 75^\circ = 15^\circ$.

$\cos 75^\circ = \cos (90^\circ - 15^\circ) = \sin 15^\circ$

The angle $15^\circ$ is between $0^\circ$ and $45^\circ$.

Now, substitute these expressions back into the original sum:

$\sin 67^\circ + \cos 75^\circ = \cos 23^\circ + \sin 15^\circ$

The expression $\sin 67^\circ + \cos 75^\circ$ is expressed in terms of trigonometric ratios of angles $23^\circ$ and $15^\circ$, both of which are between $0^\circ$ and $45^\circ$.

The expression is $\cos 23^\circ + \sin 15^\circ$.

Given:

Trigonometric ratios $\cos A$, $\tan A$, and $\sec A$.

---

To Express:

$\cos A$, $\tan A$, and $\sec A$ in terms of $\sin A$.

---

Solution:

We use fundamental trigonometric identities.

Expressing $\cos A$ in terms of $\sin A$:

We use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$.

Subtract $\sin^2 A$ from both sides:

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

Take the square root of both sides:

$\cos A = \pm \sqrt{1 - \sin^2 A}$

In problems involving angles in a right triangle, A is typically an acute angle ($0^\circ < A < 90^\circ$). In this range, $\cos A$ is positive. If A could be other angles, the sign would depend on the quadrant of A.

Assuming A is an acute angle:

$\cos A = \sqrt{1 - \sin^2 A}$

---

Expressing $\tan A$ in terms of $\sin A$:

We use the definition of tangent in terms of sine and cosine: $\tan A = \frac{\sin A}{\cos A}$.

Now substitute the expression for $\cos A$ in terms of $\sin A$ (assuming A is acute):

$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$

---

Expressing $\sec A$ in terms of $\sin A$:

We use the reciprocal identity: $\sec A = \frac{1}{\cos A}$.

Now substitute the expression for $\cos A$ in terms of $\sin A$ (assuming A is acute):

$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

Summary of the expressions in terms of $\sin A$ (assuming A is an acute angle):

$\cos A = \sqrt{1 - \sin^2 A}$

$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$

$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

Given:

The identity $\sec A (1 – \sin A)(\sec A + \tan A) = 1$.

---

To Prove:

$\sec A (1 – \sin A)(\sec A + \tan A) = 1$

---

Proof:

Consider the Left Hand Side (LHS) of the identity:

LHS = $\sec A (1 – \sin A)(\sec A + \tan A)$

Express $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$ using the identities $\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$.

Substitute these into the LHS:

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

Combine the terms within the second parenthesis:

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

Multiply the terms in the expression:

LHS = $\frac{1 \times (1 – \sin A) \times (1 + \sin A)}{\cos A \times \cos A}$

LHS = $\frac{(1 – \sin A)(1 + \sin A)}{\cos^2 A}$

Apply the algebraic identity $(a-b)(a+b) = a^2 - b^2$ to the numerator:

LHS = $\frac{1^2 – \sin^2 A}{\cos^2 A}$

LHS = $\frac{1 – \sin^2 A}{\cos^2 A}$

Use the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$, which implies $1 – \sin^2 A = \cos^2 A$.

Substitute $1 – \sin^2 A$ with $\cos^2 A$ in the numerator:

LHS = $\frac{\cos^2 A}{\cos^2 A}$

Assuming $\cos^2 A \neq 0$ (which is true for values of A where $\sec A$ and $\tan A$ are defined, i.e., $A \neq 90^\circ, 270^\circ, ...$), we can cancel the $\cos^2 A$ terms:

LHS = 1

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

To Prove

$\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$

---

Proof

We will start with the Left Hand Side (LHS) of the equation and simplify it to obtain the Right Hand Side (RHS).

LHS = $\frac{\cot A - \cos A}{\cot A + \cos A}$

We know the trigonometric identity $\cot A = \frac{\cos A}{\sin A}$. Substitute this into the LHS:

LHS = $\frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A}$

Now, take $\cos A$ as a common factor from both the numerator and the denominator:

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

Cancel the common term $\cos A$ from the numerator and the denominator (assuming $\cos A \neq 0$):

LHS = $\frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1}$

We also know the reciprocal identity $\text{cosec } A = \frac{1}{\sin A}$. Substitute this into the expression:

LHS = $\frac{\text{cosec } A - 1}{\text{cosec } A + 1}$

This expression is the same as the Right Hand Side (RHS).

Thus, LHS = RHS.

Hence, $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$.

Hence Proved.

To Prove

$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$

---

Proof

We will start by simplifying the Left Hand Side (LHS) of the equation.

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

To use the identity involving $\sec \theta$ and $\tan \theta$, we should first express the LHS in terms of these ratios. We can achieve this by dividing both the numerator and the denominator by $\cos \theta$.

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

Using the identities $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$, we get:

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

We are given to use the identity $\sec^2 \theta = 1 + \tan^2 \theta$, which can be rearranged to $1 = \sec^2 \theta - \tan^2 \theta$.

Let's substitute this expression for '1' in the denominator:

LHS = $\frac{\tan \theta - 1 + \sec \theta}{(\tan \theta - \sec \theta) + (\sec^2 \theta - \tan^2 \theta)}$

Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$, we can factorize the term $(\sec^2 \theta - \tan^2 \theta)$:

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

To simplify further, we can write $(\tan \theta - \sec \theta)$ as $-(\sec \theta - \tan \theta)$:

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

Now, take $(\sec \theta - \tan \theta)$ as a common factor from the denominator:

LHS = $\frac{\sec \theta + \tan \theta - 1}{(\sec \theta - \tan \theta) [-1 + (\sec \theta + \tan \theta)]}$

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

We can now cancel the common term $(\sec \theta + \tan \theta - 1)$ from the numerator and the denominator.

LHS = $\frac{1}{\sec \theta - \tan \theta}$

This is equal to the Right Hand Side (RHS) of the equation.

Thus, LHS = RHS.

Hence Proved.

Given:

Trigonometric ratios $\sin A$, $\sec A$, and $\tan A$.

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To Express:

$\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$.

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Solution:

We use fundamental trigonometric identities.

Expressing $\tan A$ in terms of $\cot A$:

We use the reciprocal identity: $\tan A = \frac{1}{\cot A}$.

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Expressing $\sin A$ in terms of $\cot A$:

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

Taking the square root of both sides:

$\text{cosec } A = \pm \sqrt{1 + \cot^2 A}$

In this chapter, we usually consider acute angles, where trigonometric ratios are positive. So, assuming A is an acute angle:

$\text{cosec } A = \sqrt{1 + \cot^2 A}$

We also know the reciprocal identity $\sin A = \frac{1}{\text{cosec } A}$.

Substitute the expression for $\text{cosec } A$:

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

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Expressing $\sec A$ in terms of $\cot A$:

We use the identity $\sec^2 A = 1 + \tan^2 A$.

We already expressed $\tan A$ in terms of $\cot A$: $\tan A = \frac{1}{\cot A}$.

Substitute this into the identity for $\sec^2 A$:

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

$\sec^2 A = 1 + \frac{1}{\cot^2 A}$

Find a common denominator on the right side:

$\sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$

Taking the square root of both sides, and assuming A is acute (so $\sec A > 0$):

$\sec A = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\sqrt{\cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{|\cot A|}$

For an acute angle A, $\cot A$ is positive, so $|\cot A| = \cot A$.

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$

Summary of the expressions in terms of $\cot A$ (assuming A is an acute angle):

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$

$\tan A = \frac{1}{\cot A}$

Given:

Angle A.

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To Express:

All other trigonometric ratios of $\angle$ A ($\sin A$, $\cos A$, $\tan A$, $\text{cosec } A$, $\cot A$) in terms of $\sec A$.

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Solution:

We use fundamental trigonometric identities to express each ratio in terms of $\sec A$. We assume that A is an acute angle, which is standard in this context, unless otherwise specified. For acute angles, all trigonometric ratios are positive, allowing us to take the positive square root.

1. Express $\cos A$ in terms of $\sec A$:

We use the reciprocal identity:

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

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2. Express $\tan A$ in terms of $\sec A$:

We use the identity $\sec^2 A = 1 + \tan^2 A$.

Rearrange the identity to solve for $\tan^2 A$:

$\tan^2 A = \sec^2 A - 1$

Take the square root of both sides. For acute angle A, $\tan A$ is positive, so we take the positive root:

$\tan A = \sqrt{\sec^2 A - 1}$

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3. Express $\sin A$ in terms of $\sec A$:

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

Rearrange the identity to solve for $\sin^2 A$:

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

Substitute the expression for $\cos A$ in terms of $\sec A$ ($\cos A = \frac{1}{\sec A}$):

$\sin^2 A = 1 - \left(\frac{1}{\sec A}\right)^2 = 1 - \frac{1}{\sec^2 A}$

Find a common denominator on the right side:

$\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A}$

Take the square root of both sides. For acute angle A, $\sin A$ is positive, so we take the positive root:

$\sin A = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sqrt{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{|\sec A|}$

For acute angle A, $\sec A$ is positive, so $|\sec A| = \sec A$:

$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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4. Express $\text{cosec } A$ in terms of $\sec A$:

We use the reciprocal identity $\text{cosec } A = \frac{1}{\sin A}$.

Substitute the expression for $\sin A$ in terms of $\sec A$:

$\text{cosec } A = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}$

$\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

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5. Express $\cot A$ in terms of $\sec A$:

We use the reciprocal identity $\cot A = \frac{1}{\tan A}$.

Substitute the expression for $\tan A$ in terms of $\sec A$ (assuming $\tan A > 0$):

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

The other trigonometric ratios of $\angle$ A in terms of $\sec A$ (for acute angle A) are:

$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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

$\tan A = \sqrt{\sec^2 A - 1}$

$\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

(i) $\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ}$

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Solution

We will use the complementary angle identities:

$\sin(90^\circ - \theta) = \cos \theta$

$\cos(90^\circ - \theta) = \sin \theta$

And the Pythagorean identity:

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

First, let's simplify the numerator: $\sin^2 63^\circ + \sin^2 27^\circ$.

We can write $27^\circ$ as $(90^\circ - 63^\circ)$.

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

Therefore, $\sin^2 27^\circ = \cos^2 63^\circ$.

The numerator becomes: $\sin^2 63^\circ + \cos^2 63^\circ$.

Using the Pythagorean identity, $\sin^2 63^\circ + \cos^2 63^\circ = 1$.

Next, let's simplify the denominator: $\cos^2 17^\circ + \cos^2 73^\circ$.

We can write $73^\circ$ as $(90^\circ - 17^\circ)$.

So, $\cos 73^\circ = \cos(90^\circ - 17^\circ) = \sin 17^\circ$.

Therefore, $\cos^2 73^\circ = \sin^2 17^\circ$.

The denominator becomes: $\cos^2 17^\circ + \sin^2 17^\circ$.

Using the Pythagorean identity, $\cos^2 17^\circ + \sin^2 17^\circ = 1$.

Now, we can evaluate the original expression:

$\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ} = \frac{1}{1} = 1$

Thus, the value of the expression is 1.

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(ii) $\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ$

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Solution

We will use the complementary angle identities:

$\cos(90^\circ - \theta) = \sin \theta$

$\sin(90^\circ - \theta) = \cos \theta$

The expression is: $\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ$.

We can write the angles involving $65^\circ$ in terms of $25^\circ$, since $25^\circ + 65^\circ = 90^\circ$.

$65^\circ = 90^\circ - 25^\circ$.

So, $\cos 65^\circ = \cos(90^\circ - 25^\circ) = \sin 25^\circ$.

And, $\sin 65^\circ = \sin(90^\circ - 25^\circ) = \cos 25^\circ$.

Now, substitute these values back into the original expression:

$\sin 25^\circ (\sin 25^\circ) + \cos 25^\circ (\cos 25^\circ)$

This simplifies to:

$\sin^2 25^\circ + \cos^2 25^\circ$

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we get:

$\sin^2 25^\circ + \cos^2 25^\circ = 1$

Thus, the value of the expression is 1.

(i) 9 sec² A – 9 tan² A

Justification

We have the expression $9 \sec^2 A - 9 \tan^2 A$.

Taking 9 as a common factor:

$9 (\sec^2 A - \tan^2 A)$

Using the trigonometric identity $\sec^2 A = 1 + \tan^2 A$, we can rearrange it to get $\sec^2 A - \tan^2 A = 1$.

Substituting this value into our expression:

$9 \times 1 = 9$

The correct option is (B) 9.

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(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Justification

We convert the trigonometric ratios to their sine and cosine forms.

Expression = $\left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)$

Combine the terms in each bracket by taking the common denominator:

= $\left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$

Multiply the numerators and denominators:

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

The numerator is in the form of $(a+b)(a-b) = a^2 - b^2$, where $a = (\sin \theta + \cos \theta)$ and $b = 1$.

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

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

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:

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

= $\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

= 2

The correct option is (C) 2.

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(iii) (sec A + tan A) (1 – sin A)

Justification

We convert sec A and tan A to their sine and cosine forms.

Expression = $\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 - \sin A)$

= $\left(\frac{1 + \sin A}{\cos A}\right) (1 - \sin A)$

= $\frac{(1 + \sin A)(1 - \sin A)}{\cos A}$

Using the identity $(a+b)(a-b) = a^2 - b^2$ in the numerator:

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

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

Using the identity $\cos^2 A = 1 - \sin^2 A$:

= $\frac{\cos^2 A}{\cos A}$

= $\cos A$

The correct option is (D) cos A.

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(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A}$

Justification

We use the trigonometric identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \text{cosec}^2 A$.

Expression = $\frac{\sec^2 A}{\text{cosec}^2 A}$

Now, we write sec A and cosec A in terms of sin A and cos A.

= $\frac{1/\cos^2 A}{1/\sin^2 A}$

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

= $\frac{\sin^2 A}{\cos^2 A}$

= $\tan^2 A$

The correct option is (D) tan² A.

(i) (cosec θ - cot θ)² = $\frac{1 \;-\; cos \;θ}{ 1 \;+\; cos \;θ}$

Proof

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

Convert to sin and cos:

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

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

= $\frac{(1 - \cos \theta)^2}{\sin^2 \theta}$

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

= $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$

Factoring the denominator using $a^2-b^2 = (a-b)(a+b)$:

= $\frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

= $\frac{1 - \cos \theta}{1 + \cos \theta}$ = RHS

Hence Proved.

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(ii) $\frac{cos \;A}{1 \;+\; sin \;A}$ + $\frac{1 \;+\; sin \;A}{cos \;A}$ = 2 sec A

Proof

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

Taking the LCM of the denominators:

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

Expanding the numerator:

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

Using the identity $\sin^2 A + \cos^2 A = 1$:

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

= $\frac{1 + 1 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2 + 2\sin A}{(1 + \sin A)\cos A}$

Taking 2 as a common factor in the numerator:

= $\frac{2(1 + \sin A)}{(1 + \sin A)\cos A}$

= $\frac{2}{\cos A} = 2 \sec A$ = RHS

Hence Proved.

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(iii) $\frac{tan \;θ}{1 \;-\; cot \;θ}$ + $\frac{cot \;θ}{1 \;-\; tan \;θ}$ = 1 + sec θ cosec θ

Proof

LHS = $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

Convert to sin and cos:

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

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

= $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$

= $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$

Taking a common denominator:

= $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$:

= $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:

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

Splitting the fraction:

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

= $\text{cosec } \theta \sec \theta + 1$ = RHS

Hence Proved.

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(iv) $\frac{1 \;+\; sec \;A}{sec \;A}$ = $\frac{sin^2 \;A}{1 \;-\; cos \;A}$

Proof

Simplifying LHS:

LHS = $\frac{1 + \sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1$

Simplifying RHS:

RHS = $\frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$

Since LHS = RHS, the identity is proved.

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(v) $\frac{cos \;A \;-\; sin \;A \;+\; 1}{cos \;A \;+\; sin \;A \;-\; 1}$ = cosec A + cot A

Proof

LHS = $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

Divide the numerator and denominator by $\sin A$:

= $\frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A}$

Rearranging the terms:

= $\frac{(\cot A + \text{cosec } A) - 1}{(\cot A - \text{cosec } A) + 1}$

Using the identity $1 = \text{cosec}^2 A - \cot^2 A$ in the numerator:

= $\frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{(\cot A - \text{cosec } A) + 1}$

= $\frac{(\cot A + \text{cosec } A) - (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}{\cot A - \text{cosec } A + 1}$

Taking $(\text{cosec } A + \cot A)$ common in the numerator:

= $\frac{(\text{cosec } A + \cot A) [1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1}$

= $\frac{(\text{cosec } A + \cot A) [1 - \text{cosec } A + \cot A]}{1 - \text{cosec } A + \cot A}$

= $\text{cosec } A + \cot A$ = RHS

Hence Proved.

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(vi) $\sqrt{\frac{1 \;+\; sin \;A}{1 \;-\; sin \;A}}$ = sec A + tan A

Proof

LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

Multiply the numerator and denominator by $(1 + \sin A)$ inside the square root:

= $\sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$

Using the identity $1 - \sin^2 A = \cos^2 A$:

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

= $\frac{1 + \sin A}{\cos A}$

Splitting the fraction:

= $\frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$ = RHS

Hence Proved.

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(vii) $\frac{sin \;θ \;-\; 2sin^3 \;θ}{2 cos^3 \;θ \;-\; cos \;θ}$ = tan θ

Proof

LHS = $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta}$

Taking common factors from numerator and denominator:

= $\frac{\sin \theta (1 - 2\sin^2 \theta)}{\cos \theta (2\cos^2 \theta - 1)}$

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

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$ in the numerator:

= $\tan \theta \frac{1 - 2(1 - \cos^2 \theta)}{2\cos^2 \theta - 1}$

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

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

= $\tan \theta \times 1 = \tan \theta$ = RHS

Hence Proved.

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(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Proof

LHS = $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$

Expanding the squares:

= $(\sin^2 A + 2\sin A \text{cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2\cos A \sec A \ $$ + \sec^2 A)$

Using reciprocal identities $\sin A \text{cosec } A = 1$ and $\cos A \sec A = 1$:

= $(\sin^2 A + 2 + \text{cosec}^2 A) + (\cos^2 A + 2 + \sec^2 A)$

Grouping terms:

= $(\sin^2 A + \cos^2 A) + 4 + \text{cosec}^2 A + \sec^2 A$

Using Pythagorean identities $\sin^2 A + \cos^2 A = 1$, $\text{cosec}^2 A = 1 + \cot^2 A$, $\sec^2 A = 1 + \tan^2 A$:

= $1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A)$

= $7 + \tan^2 A + \cot^2 A$ = RHS

Hence Proved.

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(ix) (cosec A – sin A)(sec A – cos A) = $\frac{1}{tan \;A \;+\; cot \;A}$

Proof

Simplifying LHS:

LHS = $(\frac{1}{\sin A} - \sin A)(\frac{1}{\cos A} - \cos A)$

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

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

Simplifying RHS:

RHS = $\frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

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

Since LHS = RHS, the identity is proved.

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(x) $\left( \frac{1\;+\;tan^{2}\;A}{1\;+\;cot^{2}\;A} \right)$ = $\left(\frac{1\;-\;tan \;A}{1\;-\;cot \;A} \right)^{2}$ = tan² A

Proof

First, we prove $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$:

LHS = $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Next, we prove $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$:

LHS = $\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2$

= $\left((1 - \tan A) \times \frac{\tan A}{\tan A - 1}\right)^2$

= $\left((1 - \tan A) \times \frac{\tan A}{-(1 - \tan A)}\right)^2$

= $(-\tan A)^2 = \tan^2 A$

Since both expressions simplify to $\tan^2 A$, the complete identity is proved.