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Introduction to Trigonometric Identities | Pythagorean Identities | Quotient and Reciprocal Identities |
Using Identities to Simplify Expressions and Prove Identities |
Fundamental Trigonometric Identities
Introduction to Trigonometric Identities
A trigonometric identity is an equation involving trigonometric ratios of an angle (or angles) that is true for all permissible values of the variable(s) for which both sides of the equation are defined. In simpler terms, it's a statement of equality between two trigonometric expressions that holds true universally, whenever the expressions make sense.
The "permissible values" refer to the domain of the trigonometric functions involved. For example, identities involving $\tan \theta$ or $\sec \theta$ are not defined for angles like $90^\circ, 270^\circ$, etc. where $\cos \theta = 0$. Similarly, identities involving $\cot \theta$ or $\text{cosec} \, \theta$ are not defined for angles like $0^\circ, 180^\circ$, etc. where $\sin \theta = 0$. Identities hold true for all other values.
Trigonometric identities are fundamental tools in mathematics and its applications in physics, engineering, and other fields. They are used extensively to:
Simplify trigonometric expressions: Allowing complex expressions to be rewritten in simpler forms.
Solve trigonometric equations: Transforming equations into solvable forms.
Prove other identities or theorems: Serving as foundational relationships.
Perform calculations in calculus: For integrating or differentiating trigonometric functions.
There are many trigonometric identities, but the most basic and frequently used ones are the Reciprocal Identities, Quotient Identities, and the Pythagorean Identities.
Pythagorean Identities
These three fundamental trigonometric identities are derived directly from the Pythagorean Theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Consider a right-angled triangle ABC, with the right angle at B. Let $\angle C = \theta$.

With respect to angle $\theta$:
Opposite Side (O) = AB
Adjacent Side (A) = BC
Hypotenuse (H) = AC
According to the Pythagorean Theorem:
$(\text{AB})^2 + (\text{BC})^2 = (\text{AC})^2$
Or, using the labels O, A, H:
$O^2 + A^2 = H^2$
... (1)
We can derive three key identities by dividing this equation by $H^2$, $A^2$, or $O^2$.
1. Identity: $\sin^2 \theta + \cos^2 \theta = 1$
Derivation:
Start with the Pythagorean Theorem (Equation 1):
$O^2 + A^2 = H^2$
[Pythagorean Theorem] ... (1)
Divide every term in Equation (1) by the square of the Hypotenuse, $H^2$. Note that $H \ne 0$ for any triangle.
$ \frac{O^2}{H^2} + \frac{A^2}{H^2} = \frac{H^2}{H^2} $
Rewrite the terms using exponents:
$ \left(\frac{O}{H}\right)^2 + \left(\frac{A}{H}\right)^2 = \left(\frac{H}{H}\right)^2 $
Recall the definitions of sine and cosine in a right triangle:
$ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{O}{H} $
$ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{A}{H} $
Also, $ \frac{H}{H} = 1 $.
Substitute these into the equation:
$ (\sin \theta)^2 + (\cos \theta)^2 = (1)^2 $
Using the standard notation $\sin^2 \theta$ for $(\sin \theta)^2$ and $\cos^2 \theta$ for $(\cos \theta)^2$:
$\mathbf{\sin^2 \theta + \cos^2 \theta = 1}$
$\sin^2 \theta + \cos^2 \theta = 1$
... (I2.1)
This identity is valid for all real values of $\theta$, as sine and cosine are defined for all angles.
Variations of this identity:
$ \sin^2 \theta = 1 - \cos^2 \theta $
$ \cos^2 \theta = 1 - \sin^2 \theta $
2. Identity: $1 + \tan^2 \theta = \sec^2 \theta$
Derivation:
Start again with the Pythagorean Theorem (Equation 1):
$O^2 + A^2 = H^2$
[Pythagorean Theorem] ... (1)
Divide every term in Equation (1) by the square of the Adjacent side, $A^2$. This requires $A \ne 0$, which means the angle $\theta$ cannot be $90^\circ, 270^\circ, \dots$, i.e., $\cos \theta \ne 0$.
$ \frac{O^2}{A^2} + \frac{A^2}{A^2} = \frac{H^2}{A^2} $
Rewrite the terms using exponents:
$ \left(\frac{O}{A}\right)^2 + \left(\frac{A}{A}\right)^2 = \left(\frac{H}{A}\right)^2 $
Recall the definitions of tangent and secant:
$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{O}{A} $
$ \sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{H}{A} $
Also, $ \frac{A}{A} = 1 $.
Substitute these into the equation:
$ (\tan \theta)^2 + (1)^2 = (\sec \theta)^2 $
Using standard notation:
$\mathbf{\tan^2 \theta + 1 = \sec^2 \theta}$ or $\mathbf{1 + \tan^2 \theta = \sec^2 \theta}$
$\sec^2 \theta - \tan^2 \theta = 1$
... (I2.2)
This identity is valid for all values of $\theta$ for which $\tan \theta$ and $\sec \theta$ are defined (i.e., $\theta \ne (2n+1)90^\circ$ for integer n).
Variations of this identity:
$ \sec^2 \theta - 1 = \tan^2 \theta $
$ \sec^2 \theta - \tan^2 \theta = 1 $ (useful for factorising difference of squares: $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$)
3. Identity: $1 + \cot^2 \theta = \text{cosec}^2 \theta$
Derivation:
Start with the Pythagorean Theorem (Equation 1):
$O^2 + A^2 = H^2$
[Pythagorean Theorem] ... (1)
Divide every term in Equation (1) by the square of the Opposite side, $O^2$. This requires $O \ne 0$, which means the angle $\theta$ cannot be $0^\circ, 180^\circ, 360^\circ, \dots$, i.e., $\sin \theta \ne 0$.
$ \frac{O^2}{O^2} + \frac{A^2}{O^2} = \frac{H^2}{O^2} $
Rewrite the terms using exponents:
$ \left(\frac{O}{O}\right)^2 + \left(\frac{A}{O}\right)^2 = \left(\frac{H}{O}\right)^2 $
Recall the definitions of cotangent and cosecant:
$ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{A}{O} $
$ \text{cosec} \, \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{H}{O} $
Also, $ \frac{O}{O} = 1 $.
Substitute these into the equation:
$ (1)^2 + (\cot \theta)^2 = (\text{cosec} \, \theta)^2 $
Using standard notation:
$\mathbf{1 + \cot^2 \theta = \text{cosec}^2 \theta}$
$\text{cosec}^2 \theta - \cot^2 \theta = 1$
... (I2.3)
This identity is valid for all values of $\theta$ for which $\cot \theta$ and $\text{cosec} \, \theta$ are defined (i.e., $\theta \ne n \times 180^\circ$ for integer n).
Variations of this identity:
$ \text{cosec}^2 \theta - 1 = \cot^2 \theta $
$ \text{cosec}^2 \theta - \cot^2 \theta = 1 $ (useful for factorising difference of squares: $(\text{cosec} \, \theta - \cot \theta)(\text{cosec} \, \theta + \cot \theta) = 1$)
These three Pythagorean identities ($\sin^2 \theta + \cos^2 \theta = 1$, $1 + \tan^2 \theta = \sec^2 \theta$, and $1 + \cot^2 \theta = \text{cosec}^2 \theta$) are fundamental and must be memorised. They are the basis for proving many other trigonometric identities.
Quotient and Reciprocal Identities
In addition to the Pythagorean identities, there are other fundamental identities that are derived directly from the definitions of the trigonometric ratios in a right triangle. These include the Reciprocal Identities and the Quotient Identities.
Reciprocal Identities
These identities express the relationship between pairs of trigonometric ratios where one is the reciprocal of the other. They follow directly from the definitions based on Opposite, Adjacent, and Hypotenuse sides (O, A, H).
Sine and Cosecant:
$ \sin \theta = \frac{O}{H} $ and $ \text{cosec} \, \theta = \frac{H}{O} $
Since $\text{cosec} \, \theta$ is the ratio $\frac{H}{O}$, which is the reciprocal of $\frac{O}{H}$, we have:
$\mathbf{\text{cosec} \, \theta = \frac{1}{\sin \theta}}$
This identity is valid for all $\theta$ where $\sin \theta \ne 0$ (i.e., $\theta \ne n \times 180^\circ$). It also implies $\sin \theta = \frac{1}{\text{cosec} \, \theta}$ and $ \sin \theta \cdot \text{cosec} \, \theta = 1 $.
Cosine and Secant:
$ \cos \theta = \frac{A}{H} $ and $ \sec \theta = \frac{H}{A} $
Since $\sec \theta$ is the ratio $\frac{H}{A}$, which is the reciprocal of $\frac{A}{H}$, we have:
$\mathbf{\sec \theta = \frac{1}{\cos \theta}}$
This identity is valid for all $\theta$ where $\cos \theta \ne 0$ (i.e., $\theta \ne (2n+1)90^\circ$). It also implies $\cos \theta = \frac{1}{\sec \theta}$ and $ \cos \theta \cdot \sec \theta = 1 $.
Tangent and Cotangent:
$ \tan \theta = \frac{O}{A} $ and $ \cot \theta = \frac{A}{O} $
Since $\cot \theta$ is the ratio $\frac{A}{O}$, which is the reciprocal of $\frac{O}{A}$, we have:
$\mathbf{\cot \theta = \frac{1}{\tan \theta}}$
This identity is valid for all $\theta$ where $\tan \theta$ is defined and non-zero (i.e., $\theta \ne n \times 90^\circ$). It also implies $\tan \theta = \frac{1}{\cot \theta}$ and $ \tan \theta \cdot \cot \theta = 1 $.
Quotient Identities
These identities express the tangent and cotangent ratios in terms of the sine and cosine ratios.
Tangent in terms of Sine and Cosine:
We know that $ \tan \theta = \frac{O}{A} $. We can express O and A in terms of H using sine and cosine definitions:
From $ \sin \theta = \frac{O}{H} $, we get $ O = H \sin \theta $.
From $ \cos \theta = \frac{A}{H} $, we get $ A = H \cos \theta $.
Now substitute these into the expression for $ \tan \theta $:
$ \tan \theta = \frac{O}{A} = \frac{H \sin \theta}{H \cos \theta} $
Assuming $H \ne 0$ and $\cos \theta \ne 0$, we can cancel H from the numerator and denominator:
$ \tan \theta = \frac{\cancel{H} \sin \theta}{\cancel{H} \cos \theta} = \frac{\sin \theta}{\cos \theta} $
Thus, the quotient identity for tangent is:
$\mathbf{\tan \theta = \frac{\sin \theta}{\cos \theta}}$
This identity is valid for all $\theta$ where $\cos \theta \ne 0$ (i.e., $\theta \ne (2n+1)90^\circ$).
Cotangent in terms of Sine and Cosine:
We know that $ \cot \theta = \frac{A}{O} $. Using $O = H \sin \theta$ and $A = H \cos \theta$:
$ \cot \theta = \frac{A}{O} = \frac{H \cos \theta}{H \sin \theta} $
Assuming $H \ne 0$ and $\sin \theta \ne 0$, we can cancel H:
$ \cot \theta = \frac{\cancel{H} \cos \theta}{\cancel{H} \sin \theta} = \frac{\cos \theta}{\sin \theta} $
Thus, the quotient identity for cotangent is:
$\mathbf{\cot \theta = \frac{\cos \theta}{\sin \theta}}$
This identity is valid for all $\theta$ where $\sin \theta \ne 0$ (i.e., $\theta \ne n \times 180^\circ$).
Note that the quotient identity for $\cot \theta$ can also be obtained directly from the reciprocal identity $\cot \theta = \frac{1}{\tan \theta}$ and the quotient identity for $\tan \theta$:
$ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\left(\frac{\sin \theta}{\cos \theta}\right)} = 1 \times \frac{\cos \theta}{\sin \theta} = \frac{\cos \theta}{\sin \theta} $
These Reciprocal and Quotient identities, along with the Pythagorean identities, form the core set of fundamental trigonometric identities. Mastery of these is crucial before moving on to more complex identities and applications.
Using Identities to Simplify Expressions and Prove Identities
Trigonometric identities are powerful tools that allow us to manipulate trigonometric expressions and equations. They serve a similar purpose to algebraic identities (like $(a+b)^2 = a^2 + 2ab + b^2$) but apply specifically to trigonometric ratios. The fundamental identities we have discussed (Reciprocal, Quotient, and Pythagorean Identities) are the building blocks for simplifying more complex expressions and proving other identities.
1. Simplifying Trigonometric Expressions
Simplifying a trigonometric expression means rewriting it in a more concise, understandable, or usable form. There is often no single "most simple" form, as the goal depends on the context (e.g., preparing for differentiation, integration, or solving an equation). However, common simplification goals include reducing the number of terms, expressing the result in terms of fewer trigonometric functions (often sine and cosine), or eliminating fractions or square roots.
General Strategies for Simplifying:
Convert to Sine and Cosine: A very common and often effective strategy is to express all tangent, cotangent, secant, and cosecant terms in terms of sine and cosine using the reciprocal and quotient identities. This reduces the number of different functions you are dealing with.
Look for Pythagorean Identities: Identify terms like $\sin^2 \theta + \cos^2 \theta$ (which equals 1), $\sec^2 \theta - 1$ (which equals $\tan^2 \theta$), $\text{cosec}^2 \theta - 1$ (which equals $\cot^2 \theta$), etc., and make the appropriate substitutions.
Apply Algebraic Techniques: Use standard algebraic methods such as:
Factoring (e.g., difference of squares, perfect squares, common factors)
Combining fractions by finding a common denominator
Multiplying by a conjugate (especially with expressions involving $1 \pm \sin \theta$ or $1 \pm \cos \theta$)
Expanding expressions (though often the goal is to avoid this)
Cancel Common Factors: After factoring or manipulating, look for terms in the numerator and denominator that can be cancelled.
Work Systematically: Don't try to do too many steps at once. Simplify one part of the expression at a time.
Example 1: Simplifying a Trigonometric Expression
Example 1. Simplify the expression: $\sec A (1 - \sin A)(\sec A + \tan A)$.
Answer:
Solution:
We are asked to simplify the expression $ \sec A (1 - \sin A)(\sec A + \tan A) $. A good approach here is to convert all terms into their sine and cosine equivalents.
Recall the reciprocal identities: $ \sec A = \frac{1}{\cos A} $
Recall the quotient identity: $ \tan A = \frac{\sin A}{\cos A} $
Substitute these into the expression:
$ \sec A (1 - \sin A)(\sec A + \tan A) = \left(\frac{1}{\cos A}\right) (1 - \sin A) \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) $
Now, simplify the terms inside the last bracket by combining the fractions, as they have a common denominator:
$ = \left(\frac{1}{\cos A}\right) (1 - \sin A) \left(\frac{1 + \sin A}{\cos A}\right) $
Next, multiply the three factors. Multiply the numerators together and the denominators together:
$ = \frac{1 \times (1 - \sin A) \times (1 + \sin A)}{\cos A \times \cos A} $
$ = \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A} $
The numerator is in the form $(a-b)(a+b)$, which simplifies to $a^2 - b^2$. Here, $a=1$ and $b=\sin A$.
$ = \frac{1^2 - \sin^2 A}{\cos^2 A} $
$ = \frac{1 - \sin^2 A}{\cos^2 A} $
Now, use the fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1$ (Identity I2.1).
Rearranging this identity, we get $1 - \sin^2 A = \cos^2 A$. Substitute this into the numerator:
$ = \frac{\cos^2 A}{\cos^2 A} $
Assuming $\cos A \ne 0$, we can cancel the common term $\cos^2 A$ from the numerator and denominator:
$ = 1 $
Thus, the simplified expression is 1.
2. Proving Trigonometric Identities
Proving a trigonometric identity means demonstrating that one side of the equation can be transformed into the other side using known identities and valid algebraic steps, for all permissible values of the variable(s).
General Strategies for Proving:
Work on One Side: Start with the side of the identity that appears more complex and try to simplify it to match the other side. It is generally easier to simplify a complicated expression than to complicate a simple one.
Convert to Sine and Cosine: As in simplification, converting all terms to sine and cosine is often a helpful first step, especially if both sides seem equally complex or involve a mix of different trigonometric functions.
Look for Algebraic Opportunities: See if you can apply factoring, combining fractions, splitting fractions, multiplying by a conjugate, etc.
Look for Identity Substitutions: Keep the fundamental identities in mind and look for opportunities to substitute expressions (e.g., replace $\tan^2 \theta + 1$ with $\sec^2 \theta$, or $1$ with $\sin^2 \theta + \cos^2 \theta$).
Manipulate Both Sides (Less Common): If you get stuck working on one side, you can sometimes work on both sides independently, transforming each into the same third expression.
Do NOT Assume the Identity is True: Do not perform operations that involve moving terms across the equality sign (e.g., adding or subtracting the same term from both sides), as this assumes the equality is already true, which is what you are trying to prove. Work only on one side at a time or manipulate both sides independently.
Be Clear and Show All Steps: Each step should be justifiable by an identity or an algebraic rule. Clearly show how you get from one step to the next.
Example 2: Proving a Trigonometric Identity
Example 2. Prove the identity: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \, \text{cosec} \, \theta$.
Answer:
To Prove:
$ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \, \text{cosec} \, \theta $
Proof:
We will start with the Left-Hand Side (LHS) as it appears more complex.
LHS = $ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} $
Convert $\tan \theta$ and $\cot \theta$ into $\sin \theta$ and $\cos \theta$ using the quotient identities ($ \tan \theta = \frac{\sin \theta}{\cos \theta} $, $ \cot \theta = \frac{\cos \theta}{\sin \theta} $):
LHS = $ \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} $
Simplify the denominators by finding a common denominator within each denominator:
The first denominator: $ 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta} $
The second denominator: $ 1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta} $
Substitute these back into the LHS expression:
LHS = $ \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}} $
To divide by a fraction, multiply by its reciprocal:
LHS = $ \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta - \sin \theta} $
Multiply the fractions:
LHS = $ \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} $
Notice that the terms in the denominators are $(\sin \theta - \cos \theta)$ and $(\cos \theta - \sin \theta)$. These are negatives of each other, i.e., $ (\cos \theta - \sin \theta) = - (\sin \theta - \cos \theta) $. We can rewrite the second term's denominator to match the first one by factoring out $-1$:
LHS = $ \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (-(\sin \theta - \cos \theta))} $
LHS = $ \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} $
Now the denominators are the same. Combine the fractions over the common denominator $ \sin \theta \cos \theta (\sin \theta - \cos \theta) $:
LHS = $ \frac{\sin \theta (\sin^2 \theta) - \cos \theta (\cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} $
LHS = $ \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} $
Use the algebraic identity for the difference of cubes in the numerator: $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $. Here, $a = \sin \theta$ and $b = \cos \theta$.
LHS = $ \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} $
Assuming $ \sin \theta - \cos \theta \ne 0 $ (i.e., $\tan \theta \ne 1$), we can cancel the common factor $ (\sin \theta - \cos \theta) $ from the numerator and denominator:
LHS = $ \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta}{\sin \theta \cos \theta} $
Use the Pythagorean identity $ \sin^2 \theta + \cos^2 \theta = 1 $ (Identity I2.1) in the numerator:
LHS = $ \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} $
Split the single fraction into two fractions:
LHS = $ \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} $
Simplify the second term and rewrite the first term:
LHS = $ \frac{1}{\sin \theta} \times \frac{1}{\cos \theta} + 1 $
Use the reciprocal identities $ \frac{1}{\sin \theta} = \text{cosec} \, \theta $ and $ \frac{1}{\cos \theta} = \sec \theta $:
LHS = $ \text{cosec} \, \theta \times \sec \theta + 1 $
Rearrange the terms to match the RHS:
LHS = $ 1 + \sec \theta \, \text{cosec} \, \theta $
This is equal to the Right-Hand Side (RHS).
Therefore, LHS = RHS. The identity is Proved.
This proof is valid for all permissible values of $\theta$, i.e., where $ \sin \theta \ne 0 $, $ \cos \theta \ne 0 $, $ \tan \theta \ne 1 $, and $ \cot \theta \ne 1 $.
Note for Competitive Exams
Simplifying expressions and proving identities are crucial skills tested in competitive exams. While there's no fixed algorithm, developing intuition comes from practice. Always keep the basic identities handy. When stuck, converting to sine and cosine is often the most reliable first step. For proofs, remember you must transform one side into the other; algebraic steps must be sound, and trigonometric substitutions must use valid identities. Be mindful of angles where ratios are undefined.
For problems involving $\sec \theta \pm \tan \theta$ or $\text{cosec} \, \theta \pm \cot \theta$, remember the variations of Pythagorean identities: $ \sec^2 \theta - \tan^2 \theta = 1 $ and $ \text{cosec}^2 \theta - \cot^2 \theta = 1 $. These can be factored as differences of squares, e.g., $ (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 $, which implies $ (\sec \theta - \tan \theta) = \frac{1}{\sec \theta + \tan \theta} $ and vice versa. This is a common trick in proofs.