CBSE Class 10 Mathematics Trigonometry Notes Set 13

Solved Examples on Trigonometric Identities

Example 1. Prove that \( \left\{ \frac{1 + \cos \theta + \sin \theta}{1 + \sin \theta - \cos \theta} \right\}^2 = \frac{1 + \cos \theta}{1 - \cos \theta} \), given that \( 1 - \cos \theta \neq 0 \).
Answer:
LHS \( = \left\{ \frac{1 + \cos \theta + \sin \theta}{1 + \sin \theta - \cos \theta} \right\}^2 \). Dividing the numerator and the denominator by \( \sin \theta \):
\( = \left\{ \frac{\operatorname{cosec} \theta + \cot \theta + 1}{\operatorname{cosec} \theta + 1 - \cot \theta} \right\}^2 \)
Using \( 1 = \operatorname{cosec}^2 \theta - \cot^2 \theta \) in the numerator: \( = \left\{ \frac{(\operatorname{cosec} \theta + \cot \theta) + (\operatorname{cosec}^2 \theta - \cot^2 \theta)}{\operatorname{cosec} \theta + 1 - \cot \theta} \right\}^2 \)
Using \( a^2 - b^2 = (a + b)(a - b) \): \( = \left\{ \frac{(\operatorname{cosec} \theta + \cot \theta) + (\operatorname{cosec} \theta + \cot \theta)(\operatorname{cosec} \theta - \cot \theta)}{\operatorname{cosec} \theta + 1 - \cot \theta} \right\}^2 = \left\{ \frac{(\operatorname{cosec} \theta + \cot \theta)(1 + \operatorname{cosec} \theta - \cot \theta)}{1 + \operatorname{cosec} \theta - \cot \theta} \right\}^2 \)
\( = (\operatorname{cosec} \theta + \cot \theta)^2 = \left( \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1 + \cos \theta}{\sin \theta} \right)^2 = \frac{(1 + \cos \theta)^2}{\sin^2 \theta} \)
\( = \frac{(1 + \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 + \cos \theta)(1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 + \cos \theta}{1 - \cos \theta} = \) RHS. Hence proved.

Example 2. Prove that \( (\sec \theta - \operatorname{cosec} \theta)(1 + \tan \theta + \cot \theta) = \sec \theta \tan \theta - \operatorname{cosec} \theta \cot \theta \).
Answer:
LHS \( = \left( \frac{1}{\cos \theta} - \frac{1}{\sin \theta} \right)\left( 1 + \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right) = \left( \frac{\sin \theta - \cos \theta}{\sin \theta \cos \theta} \right)\left( \frac{\sin \theta \cos \theta + \sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right) \)
\( = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin^2 \theta \cos^2 \theta} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin^2 \theta \cos^2 \theta} \) [using \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)]
\( = \frac{\sin^3 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{\cos^3 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{\sin \theta}{\cos^2 \theta} - \frac{\cos \theta}{\sin^2 \theta} = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} \)
\( = \tan \theta \sec \theta - \cot \theta \operatorname{cosec} \theta = \) RHS. Hence proved.

Example 3. Prove that \( \sec^6 \theta = 1 + 3\tan^2 \theta \sec^2 \theta + \tan^6 \theta \).
Answer:
LHS \( = \sec^6 \theta = (\sec^2 \theta)^3 = (1 + \tan^2 \theta)^3 \)
\( = 1^3 + (\tan^2 \theta)^3 + 3\tan^2 \theta (1 + \tan^2 \theta) \) [using \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \)]
\( = 1 + \tan^6 \theta + 3\tan^2 \theta \sec^2 \theta \) [since \( 1 + \tan^2 \theta = \sec^2 \theta \)] \( = 1 + 3\tan^2 \theta \sec^2 \theta + \tan^6 \theta = \) RHS. Hence proved.

Example 4. Prove that \( \frac{1 + \cos \theta - \sin \theta}{1 + \sin \theta + \cos \theta} + \frac{1 + \sin \theta + \cos \theta}{1 + \cos \theta - \sin \theta} = 2 \sec \theta \).
Answer:
Taking the LCM: LHS \( = \frac{(1 + \cos \theta - \sin \theta)^2 + (1 + \sin \theta + \cos \theta)^2}{[(1 + \cos \theta) + \sin \theta][(1 + \cos \theta) - \sin \theta]} \)
The numerator is \( [(1 + \cos \theta)^2 - 2\sin \theta(1 + \cos \theta) + \sin^2 \theta] + [(1 + \cos \theta)^2 + 2\sin \theta(1 + \cos \theta) + \sin^2 \theta] = 2(1 + \cos \theta)^2 + 2\sin^2 \theta = 2 + 4\cos \theta + 2(\cos^2 \theta + \sin^2 \theta) = 4 + 4\cos \theta = 4(1 + \cos \theta) \).
The denominator is \( (1 + \cos \theta)^2 - \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta + 2\cos \theta = 2\cos \theta(\cos \theta + 1) \) [using \( 1 - \sin^2 \theta = \cos^2 \theta \)].
\( \therefore \) LHS \( = \frac{4(1 + \cos \theta)}{2\cos \theta(1 + \cos \theta)} = \frac{2}{\cos \theta} = 2\sec \theta = \) RHS. Hence proved.