Grade - 9
Trigonometric Ratios
First - Part
Simplify:
1. (a) \( -\sin A ( \cos A - \sin A ) - \cos A ( \sin A - \cos A ) \)
Solution: \begin{align} & \ \ \ -\sin A ( \cos A - \sin A ) - \cos A ( \sin A - \cos A )\\ & = -\sin A \cos A + \sin^2 A-\sin A \cos A + \cos^2 A \\ & = \sin^2 A + \cos^2 A -2\sin A \cos A\\ & = 1- 2\sin A \cos A \end{align}
1. (b) \( \tan A ( \sec A +1) + \sec A ( 1-\tan A) \)
Solution: \begin{align} & \ \ \ \tan A ( \sec A +1) + \sec A ( 1-\tan A)\\ & = \tan A \sec A + \tan A + \sec A - \tan A \sec A \\ & = \tan A + \sec A \\ \end{align}
1. (c) \( (2\tan A + 1) ( 3\tan A - 8 ) \)
Solution: \begin{align} & \ \ \ (2\tan A + 1) ( 3\tan A - 8 ) \\ & = 6\tan^2 A -16\tan A +3\tan A - 8 \\ & = 6\tan^2 A - 13 \tan A - 8 \\ \end{align}
1. (d) \( (\sin B - \cos B)(\sin B + \cos B) \)
Solution: \begin{align} & \ \ \ (\sin B - \cos B) ( \sin B + \cos B) \\ & = \sin^2 B - \cos^2 B \end{align}
1. (e) \( (1-\sec A ) ( 1+ \sec A ) ( 1+ \sec^2 A) \)
Solution: \begin{align} & \ \ \ (1-\sec A ) ( 1+ \sec A ) ( 1+ \sec^2 A) \\ & = (1^2 - \sec^2 A ) (1+\sec^2 A) \\ & = (1-\sec^2 A ) (1+\sec^2 A ) \\ & = (1)^2 - ( \sec^2 A )^2 \\ & = 1 - \sec^4 A \end{align}
1. (f) \((5-\text{cosec} \beta) (2\text{cosec} \beta + 4) \)
Solution: \begin{align} & \ \ \ (5-\text{cosec} \beta) (2\text{cosec} \beta + 4) \\ & = 10 \ \text{cosec}\beta + 20 -2 \ \text{cosec}^2 \beta - 4\ \text{cosec} \beta \\ & = 20 +6\ \text{cosec} \beta - 2 \ \text{cosec}^2 \beta \end{align}
1. (g) \( (\sin A - \cos A ) ( \sin^2 A + \sin A . \cos A + \cos^2 A ) \)
Solution: \begin{align} & \ \ \ (\sin A - \cos A ) ( \sin^2 A + \sin A . \cos A + \cos^2 A ) \\ & = (\sin A )^3 - ( \cos A )^3 \\ & = \sin^3 A -\cos^3 A \end{align}
1. (h) \( (1+\tan A) ( 1- \tan A + \tan^2 A ) \)
Solution: \begin{align} & \ \ \ (1+\tan A) ( 1- \tan A + \tan^2 A ) \\ & = (1+\tan A)(1^2 -1.\tan A + \tan^2 A ) \\ & = 1^3 + (\tan A)^3 \\ & = 1 + \tan^3 A \end{align}
1. (i) \( (2\tan C - \text{cosec } B )(3\tan C + 2 \text{ cosec } B) \)
Solution: \begin{align} & \ \ \ (2\tan C - \text{cosec } B )(3\tan C + 2 \text{ cosec } B) \\ & = 6 \tan^2 C + 4 \tan C \text{ cosec } B -3 \tan C \text{ cosec } B - 2\text{ cosec}^2 B\\ & = 6 \tan^2 C + \tan C \text{ cosec } B - 2\text{ cosec}^2 B \end{align}
1. (j) \( \sin^2 x ( \sin^2 x -1) -3(\sin^4 x -3) \)
Solution: \begin{align} & \ \ \ \sin^2 x ( \sin^2 x -1) -3(\sin^4 x -3 ) \\ & = \sin^4 x - \sin^2 x - 3\sin^4 x + 9 \\ & = -2 \sin^4 x - \sin^2 x + 9 \end{align}
Resolve into factors.
2. (a) \( \sin^2 A - \cos^2 B \)
Solution: \begin{align} & \ \ \ \sin^2 A - \cos^2 B \\ & = ( \sin A -\cos B)(\sin A + \cos B ) \end{align}
2. (b) \( \sin^4 \alpha - 16\cos^4 \alpha \)
Solution: \begin{align} & \ \ \ \sin^4 \alpha - 16\cos^4 \alpha \\ & = (\sin^2 \alpha )^2 - (4\cos^2 \alpha )^2 \\ & = ( \sin^2 \alpha - 4 \cos^2 \alpha ) ( \sin^2 \alpha + 4 \cos^2 \alpha ) \\ & = \{ (\sin \alpha )^2 - (2\cos \alpha )^2 \} ( \sin^2 \alpha + 4\cos^2 \alpha ) \\ & = ( \sin \alpha - 2\cos \alpha ) ( \sin \alpha + 2\cos \alpha ) ( \sin^2 \alpha + 4\cos^2 \alpha ) \end{align}
2. (c) \( 2\sin^2\alpha - 8 \sin \alpha\)
Solution: \begin{align} & \ \ \ 2\sin^2\alpha - 8 \sin \alpha \\ & = 2\sin \alpha ( \sin \alpha - 4 ) \end{align}
2. (d) \( 2\cos^2 \alpha - 8 \cos^4 \alpha \)
Solution: \begin{align} & \ \ \ 2\cos^2 \alpha - 8 \cos^4 \alpha \\ & =2\cos^2 \alpha ( 1 - 4\cos^2 \alpha )\\ & = 2\cos^2 \alpha \left\{ 1^2 - (2\cos \alpha )^2 \right\}\\ & = 2\cos^2 \alpha (1-2\cos \alpha ) ( 1+2\cos \alpha ) \end{align}
2. (e) \( \sin^3 \theta - \cos^3 \theta \)
Solution: \begin{align} & \ \ \ \sin^3 \theta - \cos^3 \theta \\ & = ( \sin \theta - \cos \theta ) ( \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta ) \\ & = ( \sin \theta - \cos \theta ) ( \sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta )\\ & = ( \sin \theta - \cos \theta ) (1 + \sin \theta \cos \theta ) \end{align}
2. (f) \( \sin^6 A - 64\)
Solution: \begin{align} & \ \ \ \sin^6 A - 64 \\ & = ( \sin^3 A )^2 - 8^2 \\ & = ( \sin^3 A - 8)( \sin^3 A + 8 ) \\ & = ( \sin^3 A - 2^3 ) ( \sin^3 A + 2^3 ) \\ & = ( \sin A - 2 ) ( \sin^2 A + \sin A \times 2 + 2^2 )( \sin A + 2) ( \sin^2 A - \sin A \times 2 + 2^2 )\\ & = ( \sin A - 2 )(\sin A + 1) ( \sin^2 A + 2\sin A + 4 ) ( \sin^2 A -2 \sin A + 4 ) \end{align}
2. (g) \( 3\sin^2 A - 2\sin A - 5 \)
Solution: \begin{align} & \ \ \ 3\sin^2 A - 2\sin A - 5 \\ & = 3\sin^2 A - (5-3)\sin A - 5 \\ & = 3\sin^2 A - 5\sin A + 3\sin A - 5 \\ & = \sin A ( 3\sin A - 5 ) + 1(3\sin A - 5 ) \\ & = (3\sin A -5)(\sin A + 1 ) \end{align}
2. (h) \( 7\cos^2 B -17 \cos B + 6 \)
Solution: \begin{align} & \ \ \ 7\cos^2 B -17 \cos B + 6 \\ & = 7\cos^2 B -(14+3) \cos B + 6 \\ & = 7\cos^2 B -14 \cos B - 3 \cos B + 6 \\ & = 7 \cos B ( \cos B - 2 ) -3 ( \cos B - 2 )\\ & = ( \cos B -2) (7 \cos B - 3 ) \end{align}
2. (i) \( 5\tan^2 A - 27 \tan A + 10 \)
Solution: \begin{align} & \ \ \ 5\tan^2 A - 27 \tan A + 10 \\ & = 5\tan^2 A - (25 + 2) \tan A + 10 \\ & = 5\tan^2 A - 25 \tan A - 2\tan A + 10 \\ & = 5 \tan A ( \tan A - 5 ) - 2( \tan A - 5 ) \\ & = ( \tan A -5 ) (5\tan A - 2 ) \\ \end{align}
2. (j) \( \cos^2 A + 3\sin A - 3 \)
Solution: \begin{align} & \ \ \ \cos^2 A + 3\sin A - 3 \\ & = 1- \sin^2 A + 3\sin A - 3 \\ & = -\sin^2 A + 3\sin A - 2 \\ & = - ( \sin^2 A -3\sin A + 2 ) \\ & = - ( \sin^2 A - 2\sin A - \sin A + 2 ) \\ & = - \{ \sin A ( \sin A - 2 ) -1 ( \sin A - 2) \}\\ & = - ( \sin A -2 ) ( \sin A - 1 ) \end{align}
2. (k) \( -2\sin^2 A -\cos A -4 \)
Solution: \begin{align} & \ \ \ -2\sin^2 A -\cos A -4 \\ & = -2 ( 1 - \cos^2 A ) - \cos A - 4 \\ & = - 2 + 2\cos^2 A - \cos A -4 \\ & = 2\cos^2 A -\cos A - 6 \\ & = 2\cos^2 A - 4\cos A + 3\cos A - 6 \\ & = 2\cos A ( \cos A - 2 ) + 3( \cos A - 2 ) \\ & = (\cos A - 2 ) ( 2\cos A + 3 ) \\ \end{align}
2. (l) \( \sec^2 A -3\tan A + 1 \)
Solution: \begin{align} & \ \ \ \sec^2 A -3\tan A + 1 \\ & = 1 + \tan^2 A - 3\tan A + 1 \\ & = \tan^2 A - 3\tan A + 2 \\ & = \tan^2 A - 2 \tan A -\tan A + 2 \\ & = \tan A ( \tan A - 2 ) -1 (\tan A - 2 ) \\ & = ( \tan A - 2 ) ( \tan A - 1 ) \end{align}
3. Prove the following identities.
3. (a) \(\sin A . \cot A . \cos A = \frac{1}{\sec^2 A } \)
Solution: \begin{align} \text{LHS }& = \sin A . \cot A . \cos A \\ & = \sin A . \frac{\cos A }{ \sin A } . \cos A \\ & = \cos^2 A \\ & = \frac{1}{\sec^2 A }\\ & = \text{ RHS } \end{align}
3. (b) \( \tan A . \text{ cosec } A = \sec A \)
Solution: \begin{align} \text{LHS } & = \tan A . \text{ cosec } A \\ & = \frac{\sin A }{\cos A } \times \frac{1}{\sin A } \\ & = \frac{1}{\cos A} \\ & = \sec A \\ & = \text{ RHS} \end{align}
3. (c) \( \frac{\sin x } {\tan x} . \sec x = 1 \)
Solution: \begin{align} \text{LHS } & = \frac{\sin x } {\tan x} . \sec x \\ & = \sin x \times \cot x \times \sec x \\ & = \sin x \times \frac{\cos x}{\sin x} \times \frac{1}{\cos A} \\ & = 1\\ & = \text{ RHS} \end{align}
3. (d) \( \frac{\cos x}{\cot x }. \text{ cosec } x = 1 \)
Solution: \begin{align} \text{LHS } & = \frac{\cos x}{\cot x }. \text{ cosec } x \\ & = \cos x \times \tan x \times \text{ cosec } x \\ & = \cos x \times \frac{\sin x}{\cos x} \times \frac{1}{\sin x} \\ & = 1\\ & = \text{ RHS} \end{align}
3. (e) \( \sin^2 A .\text{ cosec }^2 A + \tan^2 A = \sec^2 A \)
Solution: \begin{align} \text{LHS } & = \sin^2 A .\text{ cosec }^2 A + \tan^2 A \\ & = \sin^2 A .\frac{1}{\sin^2 } + \tan^2 A \\ & = 1 + \tan^2 A \\ & = \text{ RHS} \end{align}
3. (f) \( \tan^2 A. \cot^2 A. \sin^2 A = 1 - \cos^2 A \)
Solution: \begin{align} \text{LHS } & = \tan^2 A. \cot^2 A. \sin^2 A \\ & = \frac{\sin^2 A}{\cos^2 A} \times \frac{\cos^2 A}{\sin^2 A }\times \sin^2 A \\ & = \sin^2 A \\ & = 1- \cos^2 A \\ & = \text{ RHS} \end{align}
3. (g) \( \sin B. \cot A . \cot B . \sin A = \cos A . \cos B \)
Solution: \begin{align} \text{LHS } & = \sin B. \cot A . \cot B . \sin A \\ & = \sin B \times \frac{\cos A }{ \sin A } \times \frac{\cos B}{\sin B} \times \sin A \\ & = \cos A \cos B \\ & = \text{ RHS } \end{align}
3. (h) \( ( 1-\sin^2 \alpha ) ( 1 + \tan^2 \alpha ) = 1 \)
Solution: \begin{align} \text{LHS } & = ( 1-\sin^2 \alpha ) ( 1 + \tan^2 \alpha ) \\ & = \cos^2 \alpha \sec^2 \alpha \\ & = \cos^2 \alpha \times \frac{1}{\cos^2 \alpha } \\ & = 1 \\ & = \text{ RHS} \end{align}
3. (i) \(\cot A . \sqrt{ 1 - \cos^2 A } = \cos A \)
Solution: \begin{align} \text{LHS } & = \cot A . \sqrt{ 1 - \cos^2 A } \\ & = \frac{\cos A}{\sin A} \times \sqrt{\sin^2 A} \\ & = \frac{\cos A}{\sin A} \times \sin A \\ & = \cos A \\ & = \text{ RHS} \end{align}
3. (j) \( \sqrt{1+\tan^2 A } . \sin A = \tan A \)
Solution: \begin{align} \text{LHS } & = \sqrt{1+\tan^2 A } . \sin A \\ & = \sqrt{\sec^2 A } \times \sin A \\ & = \sec A \times \sin A \\ & = \frac{1}{\cos A } \times \sin A \\ & = \frac{\sin A } {\cos A } \\ & = \tan A \\ & = \text{ RHS} \end{align}
Prove That:
4. (a) \( \sin^4 A + \cos^4 A + 2\sin^2 A . \cos^2 A = 1 \)Solution: \begin{align} \text{LHS } & = \sin^4 A + \cos^4 A + 2\sin^2 A . \cos^2 A \\ & = ( \sin^2 A )^2 + 2\sin^2 A \cos^2 A + ( \cos^2 A )^2 \\ & = ( \sin^2 A + \cos^2 A )^2 \\ & = (1)^2 \\ & = \text{ RHS} \end{align}
4. (b) \( \sin^4 A - \cos^4 A = 1 - 2\cos^2 A \)
Solution: \begin{align} \text{LHS } & = \sin^4 A - \cos^4 A \\ & = ( \sin^2 A )^2 - (\cos^2 A )^2 \\ & = ( \sin^2 A - \cos^2 A )( \sin^2 A + \cos^2 A ) \\ & = (1-\cos^2 A - \cos^2 A ) ( 1) \\ & = 1 -2\cos^2 A \\ & = \text{ RHS} \end{align}
4. (c) \( \sin^4 A + \cos^4 A = 1 - 2\sin^2 A . \cos^2 A \)
Solution: \begin{align} \text{LHS } & = \sin^4 A + \cos^4 A \\ & = ( \sin^2 A)^2 + 2\sin^2 A \cos^2 A + ( \cos^2 A )^2 - 2\sin^2 A \cos^2 A \\ & = ( \sin^2 A + \cos^2 A )^2 - 2\sin^2 A \cos^2 A \\ & = (1)^2 - 2\sin^2 A \cos^2 A \\ & = \text{ RHS} \end{align}
4. (d) \( \sin^4 A + \sin^2 A . \cos^2 A = \sin^2 A \)
Solution: \begin{align} \text{LHS } & = \sin^4 A + \sin^2 A . \cos^2 A \\ & = \sin^2 A ( \sin^2 A + \cos^2 A ) \\ & = \sin^2 A (1) \\ & = \text{ RHS} \end{align}
4. (e) \( \sin^2 B + \sin^2 B . \tan^2 B = \tan^2 B \)
Solution: \begin{align} \text{LHS } & = \sin^2 B + \sin^2 B . \tan^2 B \\ & = \sin^2 B( 1+ \tan^2 B) \\ & = \sin^2 B \sec^2 B \\ & = \sin^2 B \times \frac{1}{\cos^2 B} \\ & = \frac{\sin^2 B}{\cos^2 B } \\ & = \tan^2 B \\ & = \text{ RHS} \end{align}
4. (f) \( \cos^2 B + \cot^2 B \cos^2 B = \cot^2 B \)
Solution: \begin{align} \text{LHS } & = \cos^2 B + \cot^2 B \cos^2 B \\ & = \cos^2 B ( 1 + \cot^2 B ) \\ & =\cos^2 B \times \text{ cosec }^2 B \\ & = \cos^2 B \times \frac{1} {\sin^2 B } \\ & = \frac{\cos^2 B} { \sin^2 B} \\ & = \cot^2 B \\ & = \text{ RHS} \end{align}
4. (g) \( \sin^2 B + \sin^2 B . cot^2 B = 1 \)
Solution: \begin{align} \text{LHS } & = \sin^2 B + \sin^2 B . cot^2 B \\ & = \sin^2 B ( 1 + \cot^2 B ) \\ & = \sin^2 B \text{ cosec}^2 B \\ & = \sin^2 B \times \frac{1} { \sin^2 B} \\ & = 1 \\ & = \text{ RHS} \end{align}
4. (h) \( \cos^2 B + \cos^2 B. \tan^2 B = 1 \)
Solution: \begin{align} \text{LHS } & = \cos^2 B + \cos^2 B. \tan^2 B \\ & = \cos^2 B ( 1 + \tan^2 B) \\ & =\cos^2 B \times \sec^2 B \\ & = \cos^2 B \times \frac{1}{\cos^2 B} \\ & = 1 \\ & = \text{ RHS} \end{align}
4. (i) \( \text{cosec}^2 \alpha +\sec^2 \alpha = \text{cosec}^2 \alpha . \sec^2 \alpha \)
Solution: \begin{align} \text{LHS } & = \text{cosec}^2 \alpha +\sec^2 \alpha \\ & = \frac{1}{\sin^2 \alpha} + \frac{1}{cos^2 \alpha } \\ & = \frac{ \cos^2 \alpha +\sin^2 \alpha }{\sin^2 \alpha . \cos^2 \alpha } \\ & = \frac{1}{\sin^2 \alpha . \cos^2 \alpha }\\ & = \text{cosec}^2 \alpha \sec^2 \alpha \\ & = \text{ RHS} \end{align}
4. (j) \( \tan^2 \alpha + \cot^2 \alpha = \sec^2 \alpha + \text{cosec}^2 \alpha -2 \)
Solution: \begin{align} \text{LHS } & = \tan^2 \alpha + \cot^2 \alpha \\ & = \sec^2 \alpha - 1 + \text{cosec}^2 \alpha - 1\\ & = \sec^2 \alpha + \text{cosec}^2 \alpha -2 \\ & = \text{ RHS} \end{align}
4. (k) \( \sin \alpha . \cot \alpha - \cos \alpha . \tan \alpha = \cos \alpha - \sin \alpha \)
Solution: \begin{align} \text{LHS } & = \sin \alpha . \cot \alpha - \cos \alpha . \tan \alpha \\ & = \sin \alpha \times \frac{\cos \alpha } {\sin \alpha } - \cos \alpha \times \frac{\sin \alpha }{\cos \alpha } \\ & = \cos \alpha - \sin \alpha \\ & = \text{ RHS} \end{align}
4. (l) \( \sin^2 A - \cos^2 A + 1 = 2\sin^2 A \)
Solution: \begin{align} \text{LHS } & = \sin^2 A - \cos^2 A + 1 \\ & = \sin^2 A - ( 1 -\sin^2 A ) + 1 \\ & = \sin^2 A - 1 + \sin^2 A + 1 \\ & = 2\sin^2 A \\ & = \text{ RHS} \end{align}
4. (m) \( \cos^2 A - 1 + \sin^2 A = 0 \)
Solution: \begin{align} \text{LHS } & = \cos^2 A - 1 + \sin^2 A \\ & = \cos^2 A - 1 + 1 - \cos^2 A \\ & = 0 \\ & = \text{ RHS} \end{align}
4. (n) \( \tan^2 \theta - \sec^2 \theta + 1 = 0 \)
Solution: \begin{align} \text{LHS } & = \tan^2 \theta - \sec^2 \theta + 1 \\ & = \sec^2 \theta - 1 - \sec^2 \theta + 1 \\ & = 0 \\ & = \text{ RHS} \end{align}
4. (o) \( \cot^2 \theta - \text{cosec}^2 \theta + 1 = 0 \)
Solution: \begin{align} \text{LHS } & = \cot^2 \theta - \text{cosec}^2 \theta + 1 \\ & = \text{cosec}^2 \theta - 1 - \text{cosec}^2 \theta + 1 \\ & = 0 \\ & = \text{ RHS} \end{align}
4. (p) \( \sin \theta. \cos \theta ( \tan \theta + \cot \theta ) = 1 \)
Solution: \begin{align} \text{LHS } & = \sin \theta. \cos \theta ( \tan \theta + \cot \theta ) \\ & = \sin \theta. \cos \theta \left(\frac{\sin \theta} {\cos \theta } + \frac{\cos \theta } { \sin \theta } \right) \\ & = \sin \theta . \cos \theta \left( \frac{\sin^2 \theta + \cos^2 \theta } {\cos \theta \sin \theta } \right)\\ & = 1 \\ & = \text{ RHS} \end{align}
4. (q) \( \sqrt{ 1+ 2\sin \theta . \cos \theta } = \sin \theta + \cos \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{ 1+ 2\sin \theta . \cos \theta } \\ & = \sqrt{\sin^2 \theta + \cos^2 \theta + 2\sin \theta . \cos \theta } \\ & = \sqrt{( \sin \theta + \cos \theta )^2 }\\ & = \sin \theta + \cos \theta \\ & = \text{ RHS} \end{align}
4. (r) \( \sqrt{ \tan^2 \theta + 2 + \cot^2 \theta } = \tan \theta + \cot \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{ \tan^2 \theta + 2 + \cot^2 \theta } \\ & = \sqrt{\sec^2 \theta - 1 +2+ \text{cosec}^2 \theta - 1 } \\ & = \sqrt{ \sec^2 \theta + \text{cosec}^2 \theta }\\ & = \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta }}\\ & = \sqrt{\frac{\sin^2 \theta + \cos^2 \theta }{\cos^2 \theta.\sin^2 \theta }}\\ & = \sqrt{\frac{1}{\cos^2 \theta . \sin^2 \theta }}\\ & = \frac{1}{\cos \theta .\sin \theta} \\ & = \frac{\sin^2 \theta + \cos^2 \theta }{\cos \theta . \sin \theta } \\ & = \frac{\sin^2 \theta }{\cos \theta .\sin \theta} + \frac{\cos^2 \theta }{\cos \theta .\sin \theta}\\ & = \frac{\sin^2 \theta } { \cos \theta . \sin \theta } + \frac{ \cos^2 \theta } { \cos \theta. \sin \theta } \\ & = \frac{sin \theta }{\cos \theta } + \frac{\cos \theta }{\sin \theta } \\ & = \tan \theta + \cot \theta \\ & = \text{ RHS} \end{align}
4. (s) \( \tan^2 \beta - \sin^2 \beta = \tan^2 \beta . \sin^2 \beta \)
Solution: \begin{align} \text{LHS } & = \tan^2 \beta - \sin^2 \beta \\ & = \frac{\sin^2 \beta }{\cos^2 \beta } - \sin^2 \beta \\ & = \frac{\sin^2 \beta - \sin^2 \beta \cos^2 \beta}{\cos^2 \beta}\\ & = \frac{\sin^2 \beta ( 1 - \cos^2 \beta )}{\cos^2 \beta}\\ & = \frac{\sin^2 \beta . \sin^2 \beta }{\cos^2 \beta } \\ & = \frac{\sin^2 \beta}{\cos^2 \beta }\times \sin^2 \beta \\ & = \tan^2 \beta . \sin^2 \beta \\ & = \text{ RHS} \end{align}
Alternative
Solution: \begin{align} \text{LHS } & = \tan^2 \beta - \sin^2 \beta \\ & = \frac{\sin^2 \beta }{\cos^2 \beta } - \sin^2 \beta \\ & = \sin^2 \beta \times \sec^2 \beta - \sin^2 \beta \\ & = \sin^2 \beta (\sec^2 \beta - 1 ) \\ & = \sin^2 \beta \tan^2 \beta \\ & = \tan^2 \beta . \sin^2 \beta \\ & = \text{ RHS} \end{align}
4. (t) \( \cot^2 \beta - \cos^2 \beta = \cot^2 \beta . \cos^2 \beta \)
Solution: \begin{align} \text{LHS } & = \cot^2 \beta - \cos^2 \beta \\ & = \cot^2 \beta . \cos^2 \beta \\ & = \frac{\cos^2 \beta }{\sin^2 \beta } - \cos^2 \beta \\ & = \cos^2 \beta \times \text{ cosec}^2 \beta - \cos^2 \beta \\ & = \cos^2 \beta (\text{ cosec}^2 \beta - 1 ) \\ & = \cos^2 \beta \cot^2 \beta \\ & = \cot^2 \beta \cos^2 \beta \\ & = \text{ RHS} \end{align}
4. (u) \( \sec^4 \beta - \tan^4 \beta = \sec^2 \beta + \tan^2 \beta \)
Solution: \begin{align} \text{LHS } & = \sec^4 \beta - \tan^4 \beta \\ & = (\sec^2 \beta )^2 - ( \tan^2 \beta )^2 \\ & = ( \sec^2 \beta - \tan^2 \beta )(\sec^2 \beta + \tan^2 \beta ) \\ & = (1) ( \sec^2 \beta + \tan^2 \beta )\\ & = \sec^2 \beta + \tan^2 \beta \\ & = \text{ RHS} \end{align}
4. (v) \( \sec^4 \beta - \sec^2 \beta = \tan^4 \beta + \tan^2 \beta \)
Solution: \begin{align} \text{LHS } & = \sec^4 \beta - \sec^2 \beta \\ & = \sec^2 \beta ( \sec^2 \beta - 1 ) \\ & = (1+\tan^2 \beta ) (\tan^2 \beta )\\ & = \tan^2 \beta + \tan^4 \beta \\ & = \tan^4 \beta + \tan^2 \beta \\ & = \text{ RHS} \end{align}
5. (a) \( \frac{1}{ \sin^2 A } - \frac{1}{\tan^2 A } = 1 \)
Solution: \begin{align} \text{LHS } & = \frac{1}{ \sin^2 A } - \frac{1}{\tan^2 A } \\ & = \text{cosec}^2 A - \cot^2 A \\ & = 1 \\ & = \text{ RHS} \end{align}
5. (b) \( \frac{1}{\text{cosec}^2 A} + \frac{1}{\sec^2 A} = 1 \)
Solution: \begin{align} \text{LHS } & = \frac{1}{\text{cosec}^2 A} + \frac{1}{\sec^2 A} \\ & = \sin^2 A + \cos^2 A \\ & = 1 \\ & = \text{ RHS} \end{align}
5. (c) \(\frac{\sin A . \cos A }{\tan A } + \frac{\sin A . \cos A }{\cot A } = 1 \)
Solution: \begin{align} \text{LHS } & = \frac{\sin A . \cos A }{\tan A } + \frac{\sin A . \cos A }{\cot A } \\ & = \sin A . \cos A \times \cot A + \sin A . \cos A \times \tan A \\ & = \sin A . \cos A \times \frac{\cos A }{\sin A } + \sin A . \cos A \times \frac{\sin A }{\cos A } \\ & = \cos^2 A + \sin^2 A \\ & = 1 \\ & = \text{ RHS} \end{align}
5. (d) \(\frac{\cot^2 A } {1+\cot^2 A } = \cos^2 A \)
Solution: \begin{align} \text{LHS } & = \frac{\cot^2 A } {1+\cot^2 A } \\ & = \frac{\cot^2 A } { \text{cosec}^2 A } \\ & = \cot^2 A \times \sin^2 A \\ & = \frac{\cos^2 A }{\sin^2 A } \times \sin^2 A \\ & = \cos^2 A \\ & = \text{ RHS} \end{align}
5. (e) \( \frac{\tan^2A}{\tan^2 A + 1 } = \sin^2 A \)
Solution: \begin{align} \text{LHS } & = \frac{\tan^2A}{\tan^2 A + 1 } \\ & = \frac{\tan^2 A }{\sec^2 A } \\ & = \tan^2 A \times \cos^2 A \\ & = \frac{\sin^2 A }{\cos^2 A } \times \cos^2 A \\ & = \sin^2 A \\ & = \text{ RHS} \end{align}
5. (f) \( \frac{1-\cos^2 A + \sin^2 A}{1 - \sin^2 A + \cos^2 A} = \tan^2 A \)
Solution: \begin{align} \text{LHS } & = \frac{1-\cos^2 A + \sin^2 A}{1 - \sin^2 A + \cos^2 A} \\ & = \frac{ \sin^2 A + \sin^2 A }{\cos^2 A + \cos^2 A } = \tan^2 A \\ & = \frac{2\sin^2 A }{2\cos^2 A } \\ & = \frac{\sin^2 A }{\cos^2 A } \\ & = \tan^2 A \\ & = \text{ RHS} \end{align}
5. (g) \( \frac{1+\cos^2 A - \sin^2 A}{1+\sin^2 A - \cos^2 A } = \cot^2 A \)
Solution: \begin{align} \text{LHS } & = \frac{1+\cos^2 A - \sin^2 A}{1+\sin^2 A - \cos^2 A } \\ & = \frac{1- \sin^2 A+\cos^2 A }{1- \cos^2 A+\sin^2 A }\\ & = \frac{ \cos^2 A + \cos^2 A }{ \sin^2 A + \sin^2 A } \\ & = \frac{2\cos^2 A }{ 2\sin^2 A } \\ & = \frac{\cos^2 A }{\sin^2 A } \\ & = \cot^2 A \\ & = \text{ RHS} \end{align}
5. (h) \( \frac{\cot^2 \theta - \tan^2 \theta }{\cot \theta - \tan \theta }= \sec \theta. \text{cosec} \theta \)
Solution: \begin{align} \text{LHS } & = \frac{\cot^2 \theta - \tan^2 \theta }{\cot \theta - \tan \theta } \\ & = \frac{ \dfrac{\cos^2 \theta }{\sin^2 \theta } - \dfrac{\sin^2 \theta }{\cos^2 \theta } }{\dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }}\\ & = \frac{\frac{\cos^4 \theta - \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} }{\frac{\cos^2 \theta - \sin^2 \theta }{\sin \theta \cos \theta }} \\ & = \frac{\cos^4 \theta - \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} \times \frac{\sin \theta \cos \theta }{\cos^2 \theta - \sin^2 \theta }\\ & = \frac{(\cos^2 \theta )^2 - (\sin^2 \theta )^2 }{\sin \theta \cos \theta } \times \frac{1}{ \cos^2 \theta - \sin^2 \theta }\\ & = \frac{(\cos^2 \theta - \sin^2 \theta ) (\cos^2 \theta + \sin^2 \theta ) }{\sin \theta \cos \theta } \times \frac{1}{ \cos^2 \theta - \sin^2 \theta }\\ & = \frac{1}{\cos \theta \sin \theta } \\ & = \sec \theta. \text{cosec} \theta \\ & = \text{ RHS} \end{align}
5. (i) \( \frac{\sin^3 \theta + \cos^3 \theta }{\sin \theta + \cos \theta} = 1 - \sin \theta . \cos \theta \)
Solution: \begin{align} \text{LHS } & = \frac{\sin^3 \theta + \cos^3 \theta }{\sin \theta + \cos \theta} \\ & = \frac{(\sin \theta + \cos \theta ) ( \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) }{\sin \theta + \cos \theta } \\ & = \sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta \\ & = 1 - \sin \theta . \cos \theta \\ & = \text{ RHS} \end{align}
5. (j) \( \frac{\sin^3 \theta - \cos^3 \theta }{\sin \theta - \cos \theta } = 1 + \sin \theta . \cos \theta \)
Solution: \begin{align} \text{LHS } & = \frac{\sin^3 \theta - \cos^3 \theta }{\sin \theta - \cos \theta} \\ & = \frac{(\sin \theta - \cos \theta ) ( \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta) }{\sin \theta - \cos \theta } \\ & = \sin^2 \theta + \cos^2 \theta +\sin \theta \cos \theta \\ & = 1 + \sin \theta . \cos \theta \\ & = \text{ RHS} \end{align}
5. (k) \( \frac{\tan \theta + \cot \theta } { \sin \theta . \cos \theta } = \sec^2 \theta . \text{cosec}^2 \theta \)
Solution: \begin{align} \text{LHS } & = \frac{\tan \theta + \cot \theta } { \sin \theta . \cos \theta } \\ & = \frac{\dfrac{\sin \theta }{\cos \theta } + \dfrac{\cos \theta }{\sin \theta } }{ \sin \theta . \cos \theta}\\ & = \frac{\dfrac{\sin^2 \theta + \cos^2 \theta }{\cos \theta \sin \theta}}{\cos \theta \sin \theta} \\ & = \frac{\dfrac{1}{\cos \theta \sin \theta }}{\cos \theta \sin \theta } \\ & = \frac{1}{\cos^2 \theta \sin^2 \theta } \\ & =\sec^2 \theta . \text{cosec}^2 \theta \\ & = \text{ RHS} \end{align}
5. (l) \( \frac{\sec \theta + \cos \theta } { \sec \theta - \cos \theta } = \frac{1+\cos^2 \theta }{\sin^2 \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{\sec \theta + \cos \theta } { \sec \theta - \cos \theta } \\ & = \frac{ \dfrac{1}{\cos \theta } + \cos \theta }{\dfrac{1}{\cos \theta } - \cos \theta } \\ & = \frac{\dfrac{1 + \cos^2 \theta}{\cos \theta }}{\dfrac{1-\cos^2 \theta }{\cos \theta } } \\ & = \frac{1+ \cos^2 \theta } { 1-\cos^2 \theta }\\ & = \frac{1+\cos^2 \theta }{\sin^2 \theta } \\ & = \text{ RHS} \end{align}
Trigonometric Ratios
Second - Part
Class - 9
Prove the identities.
1. (a) \( \frac{1-\sin^4 \theta }{\cos^4 \theta } = 1+2\tan^2 \theta \)
Solution: \begin{align} \text{LHS } & = \frac{1-\sin^4 \theta }{\cos^4 \theta } \\ & = \frac{1^2 - (\sin^2 \theta )^2}{\cos^4\theta } \\ & = \frac{(1-\sin^2 \theta )(1+\sin^2 \theta)}{\cos^4 \theta }\\ & = \frac{\cos^2 \theta (1+\sin^2 \theta )}{\cos^4 \theta }\\ & = \frac{1+\sin^2 \theta }{\cos^2 \theta }\\ & = \frac{1}{\cos^2 \theta} +\frac{\sin^2 \theta }{\cos^2 \theta }\\ & = \sec^2 \theta + \tan^2 \theta \\ & = 1+\tan^2 \theta + \tan^2 \theta \\ & = 1+2\tan^2 \theta \\ & = \text{ RHS} \end{align}
Alternative Method
Solution:
\begin{align} \text{LHS } & = \frac{1-\sin^4 \theta }{\cos^4 \theta } \\ & = \frac{1}{\cos^4 \theta} -\frac{ \sin^4 \theta }{\cos^4 \theta} \\ & = \sec^4 \theta - \tan^4 \theta \\ & = ( \sec^2 \theta )^2 - (\tan^2 \theta )^2 \\ & = ( \sec^2 \theta - \tan^2 \theta )(\sec^2 \theta + \tan^2 \theta )\\ & = (1)(1+\tan^2 \theta + \tan^2 \theta )\\ & = 1+2\tan^2 \theta \\ & = \text{ RHS} \end{align}
1. (b) \( \frac{\cos^4 \theta }{1-\sin^4 \theta } = \frac{1-\sin^2 \theta }{1+ \sin^2 \theta } \)
Solution: \begin{align} \text{LHS } & =\frac{\cos^4 \theta }{1-\sin^4 \theta } \\ & = \frac{\cos^4 \theta }{1^2-(\sin^2\theta )^2 } \\ & = \frac{\cos^4 \theta }{(1-\sin^2 \theta)(1+\sin^2 \theta) } \\ & = \frac{\cos^4 \theta }{\cos^2 \theta (1+\sin^2 \theta) } \\ & = \frac{\cos^2 \theta }{1+\sin^2 \theta } \\ & = \frac{1-\sin^2 \theta }{1+\sin^2 \theta } \\ & = \text{ RHS} \end{align}
1. (c) \( \frac{1-\cos^4 \theta }{\sin^4 \theta } = 1 + 2 \cot^2 \theta \)
Solution: \begin{align} \text{LHS } & = \frac{1-\cos^4 \theta }{\sin^4 \theta } \\ & = \frac{1^2 - (\cos^2 \theta )^2}{\sin^4\theta } \\ & = \frac{(1-\cos^2 \theta )(1+\cos^2 \theta)}{\sin^4 \theta }\\ & = \frac{\sin^2 \theta (1+\cos^2 \theta )}{\sin^4 \theta }\\ & = \frac{1+\cos^2 \theta }{\sin^2 \theta }\\ & = \frac{1}{\cos^2 \theta} +\frac{\cos^2 \theta }{\sin^2 \theta }\\ & = \text{cosec}^2 \theta + \cot^2 \theta \\ & = 1+\cot^2 \theta + \cot^2 \theta \\ & = 1+2\cot^2 \theta \\ & = \text{ RHS} \end{align}
1. (d) \( \sqrt{\frac{1-\sin \theta}{1+\sin \theta }} = \sec \theta - \tan \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{\frac{1-\sin \theta}{1+\sin \theta }} \\ & = \sqrt{\frac{1-\sin \theta}{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta}} \\ & = \sqrt{\frac{(1-\sin \theta)^2}{1-\sin^2 \theta }}\\ & = \sqrt{\frac{(1-\sin \theta)^2}{\cos^2 \theta }}\\ & = \frac{1-\sin \theta}{\cos \theta }\\ & = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}\\ & = \sec \theta - \tan \theta \\ & = \text{ RHS} \end{align}
1. (e) \( \sqrt{\frac{1+\cos \theta }{1- \cos \theta }} = \text{cosec} \theta + \cot \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{\frac{1+\cos \theta }{1- \cos \theta }} \\ & = \sqrt{\frac{1+\cos \theta }{1- \cos \theta } \times \frac{1+\cos \theta}{1+\cos \theta}} \\ & = \sqrt{\frac{(1+\cos \theta)^2 }{1- \cos^2 \theta }} \\ & = \sqrt{\frac{(1+\cos \theta)^2 }{\sin^2 \theta }} \\ & = \frac{1+\cos \theta }{\sin \theta }\\ & = \frac{1}{\sin \theta } + \frac{\cos \theta}{\sin \theta} \\ & = \text{cosec} \theta + \cot \theta \\ & = \text{ RHS} \end{align}
1. (f) \( \sqrt{\frac{1+\sin \theta }{1-\sin \theta }} = \sec \theta + \tan \theta = \frac{1+\sin \theta }{\cos \theta } \)
Solution: \begin{align} \text{LHS } & = \sqrt{\frac{1+\sin \theta }{1-\sin \theta }} \\ & =\sqrt{\frac{1+\sin \theta }{1-\sin \theta } \times \frac{1+\sin \theta }{1 + \sin \theta}} \\ & = \sqrt{\frac{(1+\sin \theta)^2 }{1-\sin^2 \theta }} \\ & = \sqrt{\frac{(1+\sin \theta)^2 }{\cos^2 \theta }} \\ & = \frac{1+\sin \theta}{\cos \theta } \text{ (RHS) } \\ & = \frac{1}{\cos \theta } + \frac{\sin \theta}{\cos \theta }\\ & = \sec \theta + \tan \theta\\ & = \text{ Mid Term} \end{align}
1. (g) \( \sqrt{\frac{1-\cos \theta }{1+\cos \theta }} + \sqrt{\frac{1+\cos \theta }{1-\cos \theta }} = 2\text{cosec} \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{\frac{1-\cos \theta }{1+\cos \theta }} + \sqrt{\frac{1+\cos \theta }{1-\cos \theta }} \\ & =\sqrt{\frac{1-\cos \theta }{1+\cos \theta }\times \frac{1-\cos \theta }{1-\cos \theta }} + \sqrt{\frac{1+\cos \theta }{1-\cos \theta }\times \frac{1+ \cos \theta }{1 + \cos \theta }} \\ & = \sqrt{\frac{(1-\cos \theta)^2 }{1-\cos^2 \theta }} + \sqrt{\frac{(1+\cos \theta )^2}{1-\cos^2 \theta }}\\ & = \sqrt{\frac{(1-\cos \theta)^2 }{\sin^2 \theta }} + \sqrt{\frac{(1+\cos \theta )^2}{\sin^2 \theta }}\\ & = \frac{1-\cos \theta }{\sin \theta } + \frac{1+\cos \theta }{\sin \theta }\\ & = \frac{1- \cos \theta +1 +\cos \theta } {\sin \theta }\\ & = \frac{2}{\sin \theta } \\ & = 2\text{cosec} \theta \\ & = \text{ RHS} \end{align}
1. (h) \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta }} - \sqrt{\frac{1-\sin \theta}{1+\sin \theta }} = 2 \tan \theta \)
Solution: \begin{align} \text{LHS } & = \sqrt{\frac{1+\sin \theta}{1-\sin \theta }} - \sqrt{\frac{1-\sin \theta}{1+\sin \theta }} \\ & = \sqrt{\frac{1+\sin \theta}{1-\sin \theta }\times \frac{1+\sin \theta}{1+\sin \theta}} - \sqrt{\frac{1-\sin \theta}{1+\sin \theta }\times \frac{1-\sin \theta}{1-\sin \theta}} \\ & = \sqrt{\frac{(1+\sin \theta)^2}{1-\sin^2 \theta }} - \sqrt{\frac{(1-\sin \theta)^2}{1-\sin^2 \theta }} \\ & = \sqrt{\frac{(1+\sin \theta)^2}{\cos^2 \theta }} - \sqrt{\frac{(1-\sin \theta)^2}{\cos^2 \theta }} \\ & = \frac{1+\sin \theta}{\cos \theta } - \frac{1- \sin \theta}{\cos \theta } \\ & = \frac{1+\sin \theta - 1 +\sin \theta }{\cos \theta}\\ & = \frac{2\sin \theta}{\cos \theta} \\ & = 2\tan \theta \\ & = \text{ RHS} \end{align}
1. (i) \( \frac{1+\sin A}{1-\sin A} = ( \sec A + \tan A )^2 \)
Solution: \begin{align} \text{LHS } & = \frac{1+\sin A}{1-\sin A} \\ & = \frac{1+\sin A}{1-\sin A} \times \frac{1+\sin A}{1+\sin A}\\ & = \frac{(1+\sin A)^2}{1-\sin^2 A}\\ & = \frac{(1+\sin A)^2}{\cos^2 A}\\ & = \left( \frac{1+\sin A}{\cos A } \right)^2 \\ & = \left( \frac{1}{\cos A } + \frac{\sin A }{\cos A }\right)^2 \\ & = ( \sec A + \tan A )^2\\ & = \text{ RHS} \end{align}
1. (j) \( \frac{1-\cos A }{ 1+\cos A} = ( \text{cosec} A - \cot A )^2 \)
Solution: \begin{align} & = \frac{1-\cos A}{1+\cos A} \\ & = \frac{1-\cos A}{1+\cos A} \times \frac{1-\cos A}{1-\cos A}\\ & = \frac{(1-\cos A)^2}{1-\cos^2 A}\\ & = \frac{(1-\cos A)^2}{\sin^2 A}\\ & = \left( \frac{1-\cos A}{\sin A } \right)^2 \\ & = \left( \frac{1}{\sin A } - \frac{\cos A }{\sin A }\right)^2 \\ & = ( \text{cosec} A - \cot A )^2 \\ & = \text{ RHS} \end{align}
1. (k) \( \frac{1-\sin \theta }{1+ \sin \theta } = ( \tan \theta - \sec \theta )^2 \)
Solution: \begin{align} \text{LHS } & = \frac{1-\sin \theta}{1+\sin \theta} \\ & = \frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}\\ & = \frac{(1-\sin \theta)^2}{1-\sin^2 \theta}\\ & = \frac{(1-\sin \theta)^2}{\cos^2 \theta}\\ & = \left( \frac{1-\sin \theta}{\cos \theta } \right)^2 \\ & = \left( \frac{1}{\cos \theta } - \frac{\sin \theta }{\cos \theta }\right)^2 \\ & = ( \sec \theta - \tan \theta )^2\\ & = ( \tan \theta - \sec \theta )^2 \\ & = \text{ RHS} \end{align}
1. (l) \(\frac{\text{cosec} \theta - 1}{\text{cosec} \theta + 1} = \frac{1-\sin \theta }{1+ \sin \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{\text{cosec} \theta - 1}{\text{cosec} \theta + 1} \\ & = \frac{\dfrac{1}{\sin \theta } - 1}{\dfrac{1}{\sin \theta } + 1} \\ & = \frac{\dfrac{1-\sin \theta}{\sin \theta}}{\dfrac{1+\sin \theta }{\sin \theta }} \\ & = \frac{1-\sin \theta}{1+\sin \theta} \\ & = \text{ RHS} \end{align}
1. (m) \( \frac{\tan \theta + 1 }{\tan \theta - 1 } = \frac{1 + \cot \theta }{1- \cot \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{\tan \theta + 1 }{\tan \theta - 1 } \\ & = \frac{\dfrac{1}{\cot \theta} + 1 }{ \dfrac{1}{\cot \theta } - 1 } \\ & = \frac{\dfrac{1+\cot \theta}{\cot \theta}}{\dfrac{1-\cot \theta}{\cot \theta }} \\ & = \frac{1 + \cot \theta }{1- \cot \theta } \\ & = \text{ RHS} \end{align}
1. (n) \( \frac{\cos \theta - 1 }{\cos \theta + 1 } = \frac{1-\sec \theta } { 1 + \sec \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{\cos \theta - 1 }{\cos \theta + 1 } \\ & = \frac{\dfrac{1}{\sec \theta} - 1 }{ \dfrac{1}{\sec \theta } + 1 } \\ & = \frac{\dfrac{1-\sec \theta}{\sec \theta}}{\dfrac{1+\sec \theta}{\sec \theta }} \\ & = \frac{1 - \sec \theta }{1 + \sec \theta } \\ & = \text{ RHS} \end{align}
1. (o) \( \sec^4 A + \tan^4 A = 1 + 2\sec^2 A . \tan^2 A \)
Solution: \begin{align} \text{LHS } & = \sec^4 A + \tan^4 A \\ & = (\sec^2 A)^2 + (\tan^2 A)^2 \\ & = (\sec^2 A)^2 - 2\sec^2 A \tan^2 A + (\tan^2 A)^2 + 2\sec^2 A \tan^2 A \\ & = ( \sec^2 A - \tan^2 A )^2 + 2\sec^2 A \tan^2 A \\ & = (1)^2 + 2\sec^2 A \tan^2 A \\ & = 1 + 2\sec^2 A \tan^2 A \\ & = \text{ RHS} \end{align}
1. (p) \( \text{cosec} ^4 A + \cot^4 A = 1 + 2\text{cosec}^2 A . \cot^2 A \)
Solution: \begin{align} \text{LHS } & = \text{cosec} ^4 A + \cot^4 A \\ & = (\text{cosec} ^2 A)^2 + (\cot^2 A)^2 \\ & = (\text{cosec}^2 A)^2 - 2\text{cosec}^2 A \cot^2 A + (\cot^2 A)^2 + 2\text{cosec}^2 A \cot^2 A \\ & = ( \text{cosec}^2 A - \cot^2 A )^2 + 2\text{cosec}^2 A \cot^2 A \\ & = (1)^2 + 2\text{cosec}^2 A \cot^2 A \\ & = 1 + 2\text{cosec}^2 A \text{cosec}^2 A \\ & = \text{ RHS} \end{align}
1. (q) \(\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A . \tan B \)
Solution: \begin{align} \text{LHS } & = \frac{\tan A + \tan B}{\cot A + \cot B} \\ & = \frac{\tan A + \tan B}{\dfrac{1}{\tan A} + \dfrac{1}{\tan B} }\\ & = \frac{\tan A + \tan B}{\dfrac{\tan B + \tan A}{\tan A \tan B} } \\ & = (\tan A + \tan B) \times \frac{\tan A \tan B}{\tan A + \tan B} \\ & = \tan A . \tan B \\ & = \text{ RHS} \end{align}
1. (r) \( \frac{\tan A - \tan B}{\cot B - \cot A } = \tan A . \tan B \)
Solution: \begin{align} \text{LHS } & = \frac{\tan A - \tan B}{\dfrac{1}{\tan B} - \dfrac{1}{\tan A} } \\ & = \frac{\tan A - \tan B}{\dfrac{\tan A - \tan B}{\tan B \tan A} } \\ & = (\tan A - \tan B ) \times \frac{\tan B \tan A}{(\tan A - \tan B )}\\ & = \tan A \tan B \\ & = \text{ RHS} \end{align}
Prove the following:
2. (a) \( \frac{\sin A }{1- \cos A } = \frac{1+\cos A }{\sin A} \)
Solution: \begin{align} \text{LHS } & = \frac{\sin A }{1- \cos A } \\ & = \frac{\sin A }{1- \cos A } \times \frac{1+\cos A }{1+\cos A } \\ & = \frac{\sin A ( 1 + \cos A) }{1^2 - \cos^2 A }\\ & = \frac{\sin A ( 1 + \cos A) }{\sin^2 A }\\ & = \frac{1+\cos A }{\sin A } \\ & = \text{ RHS} \end{align} Alternative Process \begin{align} \text{LHS } & = \frac{\sin A }{1- \cos A } \\ & = \frac{\sin A }{1- \cos A } \times \frac{\sin A }{\sin A } \\ & = \frac{\sin^2 A }{\sin A ( 1 -\cos A ) }\\ & = \frac{1-\cos^2 A}{\sin A ( 1-\cos A)}\\ & = \frac{(1-\cos A)(1+\cos A)}{\sin A(1- \cos A)}\\ & = \frac{1+\cos A }{\sin A } \\ & = \text{ RHS} \end{align}
2. (b) \( \frac{\cos A }{1+\sin A } = \frac{1-\sin A}{\cos A } \)
Solution: \begin{align} \text{LHS } & = \frac{\cos A }{1+\sin A } \\ & = \frac{\cos A }{1+ \sin A } \times \frac{1-\sin A }{1-\sin A } \\ & = \frac{\cos A ( 1 - \sin A) }{1^2 - \sin^2 A }\\ & = \frac{\cos A ( 1 - \sin A) }{\cos^2 A }\\ & = \frac{1- \sin A }{\cos A } \\ & = \text{ RHS} \end{align}
2. (c) \( \frac{\tan A }{\sec A - 1 } = \frac{\sec A + 1}{\tan A } \)
Solution: \begin{align} \text{LHS } & = \frac{\tan A }{\sec A - 1 } \\ & = \frac{\tan A }{\sec A - 1 } \times \frac{\sec A + 1 }{\sec A + 1 } \\ & = \frac{\tan A (\sec A + 1 ) }{\sec^2 A - 1^2 } \\ & = \frac{\tan A (\sec A + 1 ) }{\sec^2 A - 1 } \\ & = \frac{\tan A (\sec A + 1) }{\tan^2 A } \\ & = \frac{\sec A + 1}{\tan A } \\ & = \text{ RHS} \end{align}
2. (d) \( \frac{\sec A - 1}{\tan A } = \frac{\tan A }{\sec A + 1 } \)
Solution: \begin{align} \text{LHS } & = \frac{\sec A - 1}{\tan A } \\ & = \frac{\sec A - 1}{\tan A } \times \frac{\tan A }{\tan A} \\ & = \frac{\tan A ( \sec A - 1 )}{ \tan^2 A } \\ & = \frac{\tan A ( \sec A - 1 )}{ \sec^2 A - 1 } \\ & = \frac{\tan A ( \sec A - 1 )}{ \sec^2 A - 1^2 } \\ & = \frac{\tan A ( \sec A - 1 )}{ (\sec A - 1)(\sec A + 1 ) } \\ & = \frac{\tan A }{\sec A + 1 }\\ & = \text{ RHS} \end{align}
2. (e) \( \frac{\cot A }{\text{cosec}A - 1 } = \frac{\text{cosec}A+1}{\cot A } \)
Solution: \begin{align} \text{LHS } & = \frac{\cot A }{\text{cosec}A - 1 } \times \frac{\text{cosec} A+ 1 }{\text{cosec}A + 1 } \\ & = \frac{\cot A ( \text{cosec} A + 1)}{\text{cosec}^2 A - 1^2 }\\ & = \frac{\cot A ( \text{cosec} A + 1)}{\text{cosec}^2 A - 1 }\\ & = \frac{\cot A ( \text{cosec} A + 1)}{\cot^2 A }\\ & = \frac{\text{cosec}A+1}{\cot A } \\ & = \text{ RHS} \end{align}
2. (f) \( \frac{\cot \theta } { 1+\text{cosec} \theta } = \frac{\text{cosec} \theta -1}{\cot \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{\cot \theta } { 1+\text{cosec} \theta }\\ & = \frac{\cot \theta } { 1+\text{cosec} \theta } \times \frac{\text{cosec} \theta - 1} { \text{cosec} \theta - 1}\\ & = \frac{\cot \theta ( \text{cosec} \theta - 1 )}{ \text{cosec}^2 \theta - 1^2 }\\ & = \frac{\cot \theta ( \text{cosec} \theta - 1) }{\text{cosec}^2 \theta - 1}\\ & = \frac{\cot \theta ( \text{cosec} \theta - 1) }{\cot^2 \theta}\\ & = \frac{\text{cosec} \theta -1}{\cot \theta } \\ & = \text{ RHS} \end{align}
2. (g) \( \frac{1}{\sec \theta - \tan \theta } = \sec \theta + \tan \theta \)
Solution: \begin{align} \text{LHS } & = \frac{1}{\sec \theta - \tan \theta } \\ & = \frac{\sec^2 \theta - \tan^2 \theta }{\sec \theta - \tan \theta } \\ & = \frac{(\sec \theta - \tan \theta ) ( \sec \theta + \tan \theta )}{(\sec \theta - \tan \theta) } \\ & = \sec \theta + \tan \theta \\ & = \text{ RHS} \end{align} Alternative \begin{align} \text{LHS } & = \frac{1}{\sec \theta - \tan \theta } \\ & = \frac{1}{\sec \theta - \tan \theta } \times \frac{\sec \theta + \tan \theta }{\sec \theta + \tan \theta } \\ & = \frac{\sec \theta + \tan \theta }{\sec^2 \theta - \tan^2 \theta} \\ & = \frac{\sec \theta + \tan \theta }{1} \\ & = \sec \theta + \tan \theta \\ & = \text{ RHS} \end{align}
2. (h) \( \frac{1}{\sec \theta + \tan \theta }= \sec \theta - \tan \theta \)
Solution: \begin{align} \text{LHS } & = \frac{1}{\sec \theta + \tan \theta } \\ & = \frac{\sec^2 \theta - \tan^2 \theta }{\sec \theta + \tan \theta } \\ & = \frac{(\sec \theta - \tan \theta ) ( \sec \theta + \tan \theta )}{(\sec \theta + \tan \theta) } \\ & = \sec \theta - \tan \theta \\ & = \text{ RHS} \end{align}
2. (i) \( \frac{1}{\text{cosec}\theta - \cot \theta } = \cot \theta + \text{cosec} \theta \)
Solution: \begin{align} \text{LHS } & = \frac{1}{\text{cosec}\theta - \cot \theta } \\ & = \frac{\sec^2 \theta - \tan^2 \theta }{\sec \theta - \tan \theta } \\ & = \frac{(\sec \theta - \tan \theta ) ( \sec \theta + \tan \theta )}{(\sec \theta - \tan \theta) } \\ & = \sec \theta + \tan \theta \\ & = \text{ RHS} \end{align}
2. (j) \( \frac{\sec \theta - 1}{\sec \theta + 1} = \frac{\sec^2 \theta - 2\sec \theta + 1}{\sec^2\theta - 1 } \)
Solution: \begin{align} \text{LHS } & = \frac{\sec \theta - 1}{\sec \theta + 1} \\ & = \frac{\sec \theta - 1}{\sec \theta + 1} \times \frac{\sec \theta - 1}{\sec \theta - 1} \\ & = \frac{(\sec \theta - 1)^2}{\sec^2 \theta - 1}\\ & = \frac{\sec^2 \theta - 2\times \sec \theta \times 1 + 1^2 }{\sec^2 \theta - 1}\\ & = \frac{\sec^2\theta - 2\sec \theta + 1}{\sec^2 \theta - 1}\\ & = \text{ RHS} \end{align}
Prove the following identities.
3. (a) \( \frac{1}{1-\sin B} - \frac{1}{1+\sin B} = 2 \tan B . \sec B \)
Solution: \begin{align} \text{LHS } & = \frac{1}{1-\sin B} - \frac{1}{1+\sin B} \\ & = \frac{1+\sin B - (1-\sin B)}{(1-\sin B)(1+\sin B) } \\ & = \frac{1+\sin B - 1 + \sin B}{1^2 - \sin^2 B}\\ & = \frac{2\sin B}{\cos^2 B}\\ & = 2\times\frac{\sin B}{\cos B}\times \frac{1}{\cos B} \\ & = 2\tan B \sec B \\ & = \text{ RHS} \end{align}
3. (b) \( \frac{1}{1-\cos B} - \frac{1}{1+\cos B} = 2\cot B . \text{cosec}B \)
Solution: \begin{align} \text{LHS } & = \frac{1}{1-\cos B} - \frac{1}{1+\cos B} \\ & =\frac{1+\cos B - ( 1 - \cos B)}{(1-\cos B) (1+\cos B)} \\ & =\frac{1+\cos B - 1 + \cos B}{1^2-\cos^2 B} \\ & =\frac{2 \cos B}{1-\cos^2 B} \\ & =\frac{2 \cos B}{\sin^2 B} \\ & = 2\times \frac{\cos B}{\sin B} \times \frac{1}{\sin B} \\ & = 2\cot B . \text{cosec}B \\ & = \text{ RHS} \end{align}
3. (c) \( \frac{\sin A }{1- \cos A} - \frac{\sin A}{1+\cos A} = 2\cot A\)
Solution: \begin{align} \text{LHS } & = \frac{\sin A }{1- \cos A} - \frac{\sin A}{1+\cos A} \\ & = \frac{\sin A(1+\cos A) - \sin A(1-\cos A) }{(1-\cos A)(1+\cos A)} \\ & = \frac{\sin A + \sin A \cos A - \sin A + \sin A \cos A}{1-\cos^2 A} \\ & = \frac{ 2\sin A \cos A }{\sin^2 A} \\ & = 2\times \frac{\cos A}{\sin A} \\ & = 2\cot A \\ & = \text{ RHS} \end{align}
3. (d) \( \frac{\cos A}{1-\sin A} + \frac{\cos A}{1+\sin A} = 2\sec A \)
Solution: \begin{align} \text{LHS } & = \frac{\cos A}{1-\sin A} + \frac{\cos A}{1+\sin A} \\ & = \frac{\cos A(1+\sin A) + \cos A(1-\sin A) }{(1-\sin A)(1+\sin A) } \\ & = \frac{\cos A + \sin A \cos A + \cos A - \sin A \cos A}{1^2 - \sin^2 A}\\ & = \frac{2\cos A}{1-\sin^2 A}\\ & = \frac{2\cos A}{\cos^2 A}\\ & = \frac{ 2}{\cos A}\\ & = 2\sec A\\ & = \text{ RHS} \end{align}
3. (e) \( \frac{\cos A}{1+\sin A} - \frac{\cos A \sin^2 A}{1+\sin A} = \cos A(1-\sin A) \)
Solution: \begin{align} \text{LHS } & = \frac{\cos A}{1+\sin A} - \frac{\cos A \sin^2 A}{1+\sin A} \\ & = \frac{\cos A - \cos A \sin^2 A}{1+\sin A} \\ & = \frac{\cos A ( 1 - \sin^2 A )}{1+\sin A} \\ & = \frac{\cos A(1^2 -\sin^2 A)}{1+\sin A} \\ & = \frac{\cos A(1-\sin A)(1+\sin A)}{1+\sin A}\\ & = \cos A(1-\sin A) \\ & = \text{ RHS} \end{align}
3. (f) \( \frac{\sin \alpha }{1+\cos \alpha} + \frac{1+\cos \alpha}{\sin \alpha} = 2\text{cosec }\alpha \)
Solution: \begin{align} \text{LHS } & = \frac{\sin \alpha }{1+\cos \alpha} + \frac{1+\cos \alpha}{\sin \alpha} \\ & = \frac{\sin^2 \alpha +(1+\cos \alpha)^2} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{\sin^2 \alpha +1+2\cos \alpha+ \cos^2 \alpha} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{\sin^2 \alpha+ \cos^2 \alpha +1+2\cos \alpha} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{1+1+2\cos \alpha} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{2+2\cos \alpha} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{2(1+\cos \alpha)} {(1+\cos \alpha)\sin \alpha} \\ & = \frac{2}{\sin \alpha}\\ & = 2 \text{cosec }\alpha \\ & = \text{ RHS} \end{align}
3. (g) \( \frac{1-\cos \alpha}{\sin \alpha} - \frac{\sin \alpha}{1+\cos \alpha} = 0 \)
Solution: \begin{align} \text{LHS } & = \frac{1-\cos \alpha}{\sin \alpha} - \frac{\sin \alpha}{1+\cos \alpha} \\ & = \frac{(1-\cos \alpha)(1+\cos \alpha) - \sin^2 \alpha} {\sin \alpha(1+\cos \alpha)} \\ & = \frac{1^2 - \cos^2 \alpha - \sin^2 \alpha} {\sin \alpha ( 1+\cos \alpha)}\\ & = \frac{1 - (\cos^2 \alpha + \sin^2 \alpha) } {\sin \alpha ( 1+\cos \alpha)}\\ & = \frac{1 -1} {\sin \alpha ( 1+\cos \alpha)}\\ & = \frac{0} {\sin \alpha ( 1+\cos \alpha)}\\ & = 0 \\ & = \text{ RHS} \end{align}
3. (h) \(\frac{1+\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{1-\sin \alpha} = \frac{2\cos \alpha}{1-\sin \alpha} \)
Solution: \begin{align} \text{LHS } & = \frac{1+\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{1-\sin \alpha} \\ & = \frac{(1+\sin \alpha)(1-\sin \alpha) +\cos^2 \alpha}{\cos \alpha(1-\sin \alpha)} \\ & = \frac{1^2-\sin^2 \alpha +\cos^2 \alpha}{\cos \alpha(1-\sin \alpha)} \\ & = \frac{1 -\sin^2 \alpha +\cos^2 \alpha}{\cos \alpha(1-\sin \alpha)} \\ & = \frac{\cos^2 \alpha +\cos^2 \alpha}{\cos \alpha(1-\sin \alpha)} \\ & = \frac{2\cos^2 \alpha}{\cos \alpha(1-\sin \alpha)} \\ & = = \frac{2\cos \alpha}{1-\sin \alpha} \\ & = \text{ RHS} \end{align}
3. (i) \( \frac{\cos^2 \alpha}{1+\sin \alpha} - \frac{\sin^2 \alpha}{1+\cos \alpha} = \cos \alpha - \sin \alpha \)
Solution: \begin{align} \text{LHS } & = \frac{\cos^2 \alpha}{1+\sin \alpha} - \frac{\sin^2 \alpha}{1+\cos \alpha} \\ & = \frac{1-\sin^2 \alpha}{1+\sin \alpha} - \frac{1-\cos^2 \alpha}{1+\cos \alpha}\\ & = \frac{(1-\sin \alpha)(1+\sin \alpha)}{1+\sin \alpha} - \frac{(1-\cos \alpha)(1+\cos \alpha)}{1+\cos \alpha}\\ & = 1-\sin \alpha - (1-\cos \alpha)\\ & = 1-\sin \alpha - 1 + \cos \alpha)\\ & = \cos \alpha - \sin \alpha \\ & = \text{ RHS} \end{align}
3. (j) \( \frac{\sin^2 \theta }{1-\cos \theta} - \frac{\cos^2 \theta}{1-\sin \theta} = \cos \theta - \sin \theta \)
Solution: \begin{align} \text{LHS } & =\frac{\sin^2 \theta }{1-\cos \theta} - \frac{\cos^2 \theta}{1-\sin \theta} \\ & = \frac{1-\cos^2 \theta }{1-\cos \theta} - \frac{1-\sin^2 \theta}{1-\sin \theta} \\ & =\frac{(1-\cos \theta)(1+\cos \theta) }{1-\cos \theta} - \frac{(1-\sin \theta)(1+\sin \theta)}{1-\sin \theta} \\ & = 1+\cos \theta - (1+\sin \theta )\\ & = 1 + \cos \theta - 1 - \sin \theta \\ & = \cos \theta -\sin \theta \\ & = \text{ RHS} \end{align}
3. (k) \( \frac{\tan y}{1-\cot y} + \frac{\cot y}{1- \tan y} = 1 + \sec y. \text{cosec }y \)
Solution: \begin{align} \text{LHS } & = \frac{\tan y}{1-\cot y} + \frac{\cot y}{1- \tan y} \\ & = \frac{\dfrac{\sin y }{\cos y}}{1-\dfrac{\cos y}{\sin y}} + \frac{\dfrac{\cos y }{\sin y}}{ 1 - \dfrac{\sin y}{\cos y}} \\ & = \frac{\dfrac{\sin y }{\cos y}}{\dfrac{\sin y - \cos y}{\sin y}} + \frac{\dfrac{\cos y }{\sin y}}{\dfrac{\cos y - \sin y}{\cos y}} \\ & = \frac{\sin y }{\cos y} \times \frac{\sin y}{\sin y - \cos y} + \frac{\cos y}{\sin y} \times \frac{\cos y}{\cos y - \sin y} \\ & = \frac{\sin^2 y }{\cos y(\sin y - \cos y)} - \frac{\cos^2 y}{\sin y(\sin y - \cos y)} \\ & = \frac{\sin^3 y - \cos^3 y}{\cos y \sin y (\sin y - \cos y)} \\ & = \frac{(\sin y - \cos y)(\sin^2 y + \sin y \cos y + \cos^2 y )}{\cos y \sin y (\sin y - \cos y)} \\ & = \frac{\sin^2 y + \cos^2 y + \sin y \cos y }{\cos y \sin y } \\ & = \frac{1 + \sin y \cos y }{\cos y \sin y } \\ & = \frac{1 }{\cos y \sin y } + \frac{\sin y \cos y}{\cos y \sin y} \\ \\ & = \sec y \text{ cosec } y + 1 \\ & = 1 + \sec y \text{ cosec } y \\ & = \text{ RHS} \end{align}
Alternative \begin{align} \text{LHS } & = \frac{\tan y}{1-\cot y} + \frac{\cot y}{1- \tan y} \\ & = \frac{\tan y}{1-\dfrac{1}{\tan y}} + \frac{1}{\tan y (1- \tan y)} \\ & = \frac{\tan y}{\dfrac{\tan y - 1}{\tan y}} + \frac{1}{\tan y (1- \tan y)} \\ & = \frac{\tan^2 y}{\tan y - 1} - \frac{1}{\tan y ( \tan y - 1 )} \\ & = \frac{\tan^3 y - 1}{\tan y ( \tan y - 1)}\\ & = \frac{\tan^3 y - 1^3}{\tan y ( \tan y - 1)}\\ & = \frac{(\tan y - 1)(\tan^2 y + \tan y \times 1 + 1^2) }{\tan y ( \tan y - 1)}\\ & = \frac{\tan^2 y + \tan y + 1 }{\tan y }\\ & = \frac{\tan^2 y}{\tan y} + \frac{\tan y }{\tan y} + \frac{1}{\tan y}\\ & = \tan y + 1 + \cot y \\ & = 1 + \tan y + \cot y \\ & = 1 + \frac{\sin y}{\cos y} + \frac{\cos y}{\sin y} \\ & = 1 + \frac{\sin^2 y + \cos^2 y }{\cos y \sin y } \\ & = 1 + \frac{1}{\cos y \sin y}\\ & = 1 + \sec y \text{ cosec } y \\ & = \text{ RHS} \end{align}
3. (l) \( \frac{\cos^2 y}{1-\tan y} + \frac{\sin^2 y}{1-\cot y} = 1 + \sin y. \cos y \)
Solution: \begin{align} \text{LHS } & = \frac{\cos^2 y}{1-\tan y} + \frac{\sin^2 y}{1-\cot y}\\ & = \frac{\cos^2 y}{1-\dfrac{\sin y}{\cos y}} + \frac{\sin^2 y}{1-\dfrac{\cos y}{\sin y} }\\ & = \frac{\cos^2 y}{\dfrac{\cos y - \sin y}{\cos y}} + \frac{\sin^2 y}{\dfrac{\sin y - \cos y}{\sin y} }\\ & = \frac{\cos^3 y}{\cos y - \sin y} + \frac{\sin^3 y }{\sin y - \cos y } \\ & = \frac{\cos^3 y}{\cos y - \sin y} - \frac{\sin^3 y }{\cos y - \sin y } \\ & = \frac{\cos^3 y - \sin^3 y }{\cos y - \sin y } \\ & = \frac{(\cos y - \sin y )(\cos^2 y + \cos y \sin y + \sin^2 y )}{\cos y - \sin y } \\ & = \sin^2 y + \cos^2 y + \sin y \cos y \\ & = 1 + \sin y \cos y \\ & = \text{ RHS} \end{align}
3. (m) \( \frac{\tan^2 A}{\cos^2 B} - \frac{\tan^2 B}{\cos^2 A} = \sec^2 A - \sec^2 B \)
Solution: \begin{align} \text{LHS } & = \frac{\tan^2 A}{\cos^2 B} - \frac{\tan^2 B}{\cos^2 A} \\ & = \tan^2 A \times \frac{1}{\cos^2 B} - \tan^2 B \times \frac{1}{\cos^2 A} \\ & = \tan^2 A \sec^2 B - \tan^2 B \sec^2 A \\ & = (\sec^2 A - 1) \sec^2 B - (\sec^2 B - 1) \sec^2 A \\ & = \sec^2 A \sec^2 B - \sec^2 B - \sec^2 A \sec^2 B + \sec^2 A \\ & = \sec^2 A - \sec^2 B \\ & = \text{ RHS} \end{align}
3. (n) \( (p\sin \beta - r\cos \beta)^2 +(r\sin \beta + p\cos \beta)^2 = p^2 + r^2 \)
Solution: \begin{align} \text{LHS } & = (p\sin \beta - r\cos \beta)^2 +(r\sin \beta + p\cos \beta)^2 = p^2 + r^2 \\ & =(p\sin \beta)^2 - 2\times p\sin \beta \times r\cos \beta + (r\cos \beta )^2 \\ & +(r\sin \beta)^2 + 2\times r\sin \beta \times p\cos \beta + (p\cos \beta )^2 \\ & = p^2\sin^2 \beta -2pr\sin \beta \cos \beta + r^2\cos^2 \beta \\ & + r^2\sin^2 \beta + 2pr\sin \beta \cos \beta + p^2 \cos^2 \beta \\ & = p^2 \sin^2 \beta + p^2\cos^2 \beta + r^2 \sin^2 \beta + r^2 \cos^2 \beta \\ & = p^2 (\sin^2 \beta + \cos^2 \beta ) + r^2 (\sin^2 \beta + \cos^2 \beta )\\ & = p^2 (1) + r^2 (1) \\ & = p^2 + r^2 \\ & = \text{ RHS} \end{align}
3. (o) \( (\sin A. \sin B)^2 + (\cos A)^2 - (\cos A. \cos B)^2 = \sin^2 B \)
Solution: \begin{align} \text{LHS } & = (\sin A. \sin B)^2 + (\cos A)^2 - (\cos A. \cos B)^2 \\ & = \sin^2A \sin^2 B + \cos^2 A - \cos^2 A \cos^2 B \\ & = \sin^2A \sin^2 B + \cos^2 A - (1-\sin^2 A)(1-\sin^2 B ) \\ & = \sin^2 A \sin^2 B + \cos^2 A - (1-\sin^2 B - \sin^2 A + \sin^2 A \sin^2 B ) \\ & = \sin^2 A \sin^2 B + \cos^2 A - 1 + \sin^2 B + \sin^2 A - \sin^2 A \sin^2 B ) \\ & = \sin^2 A +\cos^2 A - 1 + \sin^2 B \\ & = 1 - 1 + \sin^2 B \\ & = \sin^2 B \\ & = \text{ RHS} \end{align}
3. (p) \( ( \cos \alpha \cos \beta )^2 + (\sin \beta )^2 - (\sin \alpha . \sin \beta )^2 = \cos^2 \alpha \)
Solution: \begin{align} \text{LHS } & = ( \cos \alpha \cos \beta )^2 + (\sin \beta )^2 - (\sin \alpha . \sin \beta )^2 \\ & = \cos^2 \alpha \cos^2 \beta + \sin^2 \beta - \sin^2 \alpha \sin^2 \beta \\ & = \cos^2 \alpha \cos^2 \beta + \sin^2 \beta - (1-\cos^2 \alpha )(1-\cos^2 \beta ) \\ & = \cos^2 \alpha \cos^2 \beta + \sin^2 \beta - ( 1 - \cos^2 \beta - \cos^2 \alpha + \cos^2 \alpha \cos^2 \beta )\\ & = \cos^2 \alpha \cos^2 \beta + \sin^2 \beta - 1 + \cos^2 \beta + \cos^2 \alpha - \cos^2 \alpha \cos^2 \beta \\ & = \sin^2 \beta + \cos^2 \beta - 1 + \cos^2 \alpha \\ & = 1 - 1 + \cos^2 \alpha \\ & = 0 + \cos^2 \alpha \\ & = \cos^2 \alpha \\ & = \text{ RHS} \end{align}
Trigonometric Ratios
Second - Third
Class - 9
Prove the following identities:
1. (a) \( (1+\sin A + \cos A)^2 = 2(1+\sin A)(1+\cos A) \)
Solution: \begin{align} \text{LHS} & = ( 1+ \sin A + \cos A)^2 \\ \ & = [(1+\sin A ) + \cos A ]^2\\ \ & = (1+\sin A)^2 + 2\times (1+\sin A) \times \cos A + \cos^2 A \\ \ & = (1+\sin A)^2 + 2(1+\sin A) \cos A + 1-\sin^2 A \\ \ & = (1+\sin A)^2 + 2(1+\sin A)\cos A + 1^2 -\sin^2 A \\ \ & = (1+\sin A)^2 + 2(1+\sin A)\cos A + (1 + \sin A )(1-\sin A) \\ \ & = (1+\sin A) ( 1+\sin A + 2\cos A + 1-\sin A )\\ \ & = (1+\sin A) ( 2+ 2\cos A )\\ \ & = (1+\sin A) . 2(1+\cos A) \\ \ & = 2(1+\sin A)(1+\cos A) \\ \ & = \text{RHS}\\ \end{align} Alternative Method \begin{align} \text{LHS} & = ( 1+ \sin A + \cos A)^2 \\ & [ \because (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca ] \\ & = 1^2 + \sin^2 A + \cos^2 A + 2\times 1 \times \sin A + 2 \times \sin A \times \cos A + 2 \cos A \times 1 \\ & = 1 + 1 + 2\sin A + 2 \sin A \cos A + 2\cos A \\ & = 2 + 2 \sin A + 2 \sin A \cos A + 2\cos A \\ & = 2( 1 + \sin A ) + 2\cos A ( \sin A + 1 ) \\ & = (1 + \sin A ) ( 2 + 2\cos A ) \\ & = ( 1 + \sin A ). 2( 1 + \cos A ) \\ & = 2(1+\sin A) ( 1 + \cos A ) \end{align}
1. (b) \( (1-\sin A + \cos A)^2 = 2(1-\sin A)(1+\cos A) \)
Solution: \begin{align} \text{LHS} & = ( 1-\sin A + \cos A)^2 \\ \ & = [( 1-\sin A) + \cos A ]^2 \\ \ & = (1-\sin A)^2 +2\times (1-\sin A ) \times \cos A + \cos^2 A \\ \ & = (1-\sin A)^2 +2\times (1-\sin A ) \times \cos A + 1-\sin^2 A \\ \ & = (1-\sin A)^2 +2\times (1-\sin A ) \times \cos A + (1-\sin A)(1+\sin A ) \\ \ & = (1-\sin A) (1-\sin A +2\cos A + 1+\sin A )\\ \ & = (1-\sin A) ( 2+2\cos A )\\ \ & = (1-\sin A)\times 2 ( 1 + \cos A )\\ \ & = 2(1-\sin A) ( 1 + \cos A )\\ \ & = \text{RHS}\\ \end{align} Alternative Method \begin{align} \text{LHS} & = ( 1- \sin A + \cos A)^2 \\ & [ \because (a-b+c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca ] \\ & = 1^2 + \sin^2 A + \cos^2 A - 2\times 1 \times \sin A - 2 \times \sin A \times \cos A + 2 \cos A \times 1 \\ & = 1 + 1 - 2\sin A - 2 \sin A \cos A + 2\cos A \\ & = 2 + 2\cos A - 2 \sin A - 2 \sin A \cos A \\ & = 2( 1 + \cos A ) - 2\sin A ( 1+\cos A ) \\ & = (2 - 2\sin A ) ( 1+ \cos A ) \\ & = 2 ( 1 - \sin A ) ( 1 + \cos A ) \\ & = 2(1-\sin A) ( 1 + \cos A ) \end{align}
1. (c) \((1+\sin A )^2 - (1-\sin A)^2 = 4\sin A\)
Solution: \begin{align} \text{LHS} & = (1+\sin A )^2 - (1-\sin A)^2 \\ & = [1+\sin A + 1 - \sin A ] [ 1 + \sin A - ( 1 - \sin A ) ] \\ & = 2 ( 1 + \sin A - 1 + \sin A ) \\ & = 2 ( 2\sin A ) \\ & = 4 \sin A \end{align}
1. (d) \( (\sin \beta + \cos \beta )^3 = 3 (\sin \beta + \cos \beta ) - 2(\sin^3 \beta + \cos^3 \beta ) \)
Solution: \begin{align} \text{LHS} & = (\sin \beta + \cos \beta )^3 \\ & [\because (a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 ]\\ & = \sin^3 \beta + 3\sin^2 \beta \cos \beta + 3\sin \beta \cos^2 \beta + \cos^3 \beta \\ & = \sin^3 \beta + 3( 1 - \cos^2 \beta ) \cos \beta + 3 \sin \beta ( 1 - \sin^2 \beta ) + \cos^3 \beta \\ & = \sin^3 \beta + 3\cos \beta - 3 \cos^3 \beta + 3\sin \beta - 3\sin^3 \beta + \cos^3 \beta \\ & = 3\sin \beta + 3\cos \beta - 2\sin^3 \beta - 2\cos^3 \beta \\ & = 3( \sin \beta + \cos \beta ) - 2(\sin^3 \beta + \cos^3 \beta ) \\ & = \text{RHS} \end{align}
1. (e) \( \cos^8 \theta - \sin^8 \theta = (1-2\sin^2 \theta )(1-2\sin^2 \theta + 2\sin^4 \theta ) \)
Solution: \begin{align} \text{LHS} & = \cos^8 \theta - \sin^8 \theta \\ & = ( \cos^4 \theta )^2 - ( \sin^4 \theta )^2 \\ & = ( \cos^4 \theta - \sin ^4 \theta ) ( \cos^4 \theta + \sin^4 \theta ) \\ & = [ ( \cos^2 \theta )^2 - ( \sin^2 \theta )^2 ] [ ( \cos^2 \theta )^2 + \sin^4 \theta ] \\ & = ( \cos^2 \theta -\sin^2 \theta ) ( \cos^2 \theta + \sin^2 \theta ) [ ( 1-\sin^2 \theta )^2 + \sin^4 \theta ] \\ & = ( 1-\sin^2 \theta -\sin^2 \theta ) ( 1) [ 1^2 - 2 \times 1 \times \sin^2 \theta + ( \sin^2 \theta )^2 + \sin^4 \theta ] \\ & = ( 1-2\sin^2 \theta )[ 1^2 - 2\sin^2 \theta + \sin^4 \theta + \sin^4 \theta ]\\ & = (1-2\sin^2 \theta )(1-2\sin^2 \theta + 2\sin^4 \theta ) \\ & = \text{ RHS} \end{align}
1. (f) \( \sin^6 \delta - \cos^6 \delta = (1-2\cos^2 \delta)(\cos^2 \delta + \sin^4 \delta ) \)
Solution: \begin{align} \text{LHS} & = \sin^6 \delta - \cos^6 \delta \\ & = ( \sin^2 \delta )^3 - (\cos^2 \delta )^3 \\ & = (\sin^2 \delta - \cos^2 \delta )[ (\sin^2 \delta)^2 + \sin^2 \delta \cos^2 \delta + (\cos^2 \delta)^2 ]\\ & = (1-\cos^2 \delta - \cos^2 \delta )(\sin^4 \delta + \sin^2 \delta \cos^2 \delta + \cos^4 \delta )\\ & = (1- 2\cos^2 \delta )(\sin^4 \delta + \cos^2 \delta ( \sin^2 \delta + \cos^2 \delta )]\\ & = (1- 2\cos^2 \delta )(\sin^4 \delta + \cos^2 \delta ( 1)]\\ & = (1-2\cos^2 \delta)(\cos^2 \delta + \sin^4 \delta ) \\ & = \text{ RHS} \end{align}
1. (g) \( \sec^6 \delta - \cos^6 \delta = (\sin^2 \delta + \tan^2 \delta )(1+\cos^4 \delta + \sec^4 \delta ) \)
Solution: \begin{align} \text{LHS} & = \sec^6 \delta - \cos^6 \delta \\ & = (\sec^2 \delta )^3 - ( \cos^2 \delta )^3 \\ & = ( \sec^2 \delta - \cos^2 \delta ) [ (\sec^2 \delta )^2 + \sec^2 \delta \cos^2 \delta + ( \cos^2 \delta )^2 ] \\ & = [( 1+ \tan^2 \delta - (1-\sin^2 \delta )] ( \sec^4 \delta + 1 + \cos^4 \delta ) \\ & = (1+\tan^2 \delta - 1 + \sin^2 \delta ) ( \sec^4 \delta + 1 + \cos^4 \delta ) \\ & = (\sin^2 \delta + \tan^2 \delta )(1+\cos^4 \delta + \sec^4 \delta ) \\ & = \text{ RHS} \end{align}
1. (h) \( \frac{ \sin A - 2\sin^3 A }{2\cos^3 A - \cos A } = \tan A \)
Solution: \begin{align} \text{LHS} & =\frac{ \sin A - 2\sin^3 A }{2\cos^3 A - \cos A } \\ & = \frac{\sin A ( 1 - 2 \sin^2 A )}{ \cos A ( 2\cos^2 A - 1) } \\ & = \frac{\sin A ( 1 - 2 \sin^2 A )}{ \cos A [ 2(1 -\sin^2 A )- 1] } \\ & = \frac{\sin A ( 1 - 2 \sin^2 A )}{ \cos A ( 2 - 2\sin^2 A - 1 ) } \\ & = \frac{\sin A ( 1 - 2 \sin^2 A )}{ \cos A ( 1 - 2\sin^2 A ) } \\ & = \frac{\sin A}{ \cos A } \\ & = \tan A \\ & = \text{ RHS} \end{align}
1. (i) \( \frac{\sec A + \tan A }{\sec A + \tan A - \cos A } = \text{cosec } A \)
Solution: \begin{align} \text{LHS} & = \frac{\sec A + \tan A }{\sec A + \tan A - \cos A } \\ & = \frac{\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A } } {\dfrac{ 1}{ \cos A} +\dfrac{\sin A}{ \cos A } -\cos A }\\ & = \frac{\dfrac{1+ \sin A }{ \cos A} }{\dfrac{1+\sin A - \cos^2 A }{ \cos A } } \\ & = \frac{1+\sin A }{ 1+\sin A - (1-\sin^2 A )}\\ & = \frac{1+\sin A }{ (1+\sin A)(1 - 1 + \sin A )}\\ & = \frac{1+\sin A }{ (1+\sin A)(\sin A )}\\ & = \frac{1}{\sin A } \\ & = \text{cosec } A \\ & = \text{ RHS} \end{align}
1. (j) \( \sin^6 A + \cos^6 A = 1 -3\sin^2 A + 3 \sin^4 A \)
Solution: \begin{align} \text{LHS} & = \sin^6 A + \cos^6 A \\ & = ( \sin^2 A )^3 + ( \cos^2 A )^3 \\ & = (\sin^2 A + \cos^2 A )[(\sin^2 A )^2 - \sin^2 A \cos^2 A + (\cos^2 A )^2 ] \\ & = (1) [ \sin^4 A - \sin^2 A (1-\sin^2 A) + (1-\sin^2 )^2 ]\\ & = \sin^4 A -\sin^2 A + \sin^4 A + 1^2 - 2\times 1 \times \sin^2 A + (\sin^2 )^2 A ] \\ & = 2\sin^4 A - \sin^2 A + 1 - 2\sin^2 A + \sin^4 A\\ & = 1 -3\sin^2 A + 3 \sin^4 A\\ & = \text{ RHS} \end{align}
1. (k) \( (1-\tan \gamma )^2 + (1-\cot \gamma )^2 = (\text{cosec } \gamma - \sec \gamma )^2 \)
Solution: \begin{align} \text{LHS} & = (1-\tan \gamma )^2 + (1-\cot \gamma )^2 \\ & = 1^2 - 2\times 1 \times \tan \gamma + \tan^2 \gamma + 1^2 - 2\times \cot \gamma + \cot^2 \gamma \\ & = 1 - 2\tan \gamma + \tan^2 \gamma + 1 - 2\cot \gamma + \cot^2 \gamma\\ & = 1+\cot^2 \gamma - 2\tan \gamma - 2\cot \gamma + 1 + \tan^2 \gamma \\ & = \text{cosec}^2 \gamma - 2 (\tan \gamma + \cot \gamma ) + \sec^2 \gamma \\ & = \text{cosec}^2 \gamma - 2 \left(\frac{\sin \gamma }{\cos \gamma } + \frac{\cos \gamma}{\sin \gamma} \right) + \sec^2 \gamma \\ & = \text{cosec}^2 \gamma - 2 \left(\frac{\sin^2 \gamma +\cos^2 \gamma}{\cos \gamma \sin \gamma} \right) + \sec^2 \gamma \\ & = \text{cosec}^2 \gamma - 2 \left(\frac{1}{\cos \gamma \sin \gamma} \right) + \sec^2 \gamma \\ & = \text{cosec}^2 \gamma - 2\sec \gamma \text{cosec } \gamma + \sec^2 \gamma \\ & = (\text{cosec } \gamma - \sec \gamma )^2 \\ & = \text{ RHS} \end{align}
1. (l) \( \frac{\sec^6 A - \tan^6 A }{\sec^2 A - \tan^2 A } = 1 + 3\sec^2 A . \tan^2 A \)
Solution: \begin{align} \text{LHS }& = \frac{\sec^6 A - \tan^6 A }{\sec^2 A - \tan^2 A } \\ & = \frac{(\sec^2 A)^3 - (\tan^2 A)^3 }{\sec^2 A - \tan^2 A }\\ & = \frac{(\sec^2 A - \tan^2 A ) \{(\sec^2 A)^2 + \sec^2 A \tan^2 A + (\tan^2 A )^2 \} }{\sec^2 A - \tan^2 A }\\ & = (\sec^2 A)^2 + \sec^2 A \tan^2 A + (\tan^2 A )^2 \\ & = (\sec^2 A)^2 -2 \sec^2 A \tan^2 A + (\tan^2 A )^2 + 3\sec^2 A \tan^2 A \\ & = (\sec^2 A -\tan^2 A )^2 + 3\sec^2 A \tan^2 A \\ & = (1)^2 + 3\sec^2 A \tan^2 A \\ & = 1 + 3\sec^2 A \tan^2 A \\ & = \text{ RHS} \end{align}
1. (m) \( \frac{\cos A + \cos B}{\sin A - \sin B} = \frac{ \sin B + \sin A }{\cos B - \cos A } \)
Solution: \begin{align} \text{LHS } & = \frac{\cos A + \cos B}{\sin A - \sin B} \\ & = \frac{\cos A + \cos B}{\sin A - \sin B} \times \frac{\cos A - \cos B}{\cos A - \cos B} \times \frac{ \sin A + \sin B}{\sin A + \sin B} \\ & = \frac{ ( \cos A + \cos B ) (\cos A - \cos B)}{ (\sin A - \sin B)(\sin A - \sin B) } \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{\cos^2 A - \cos^2 B}{\sin^2 A - \sin^2 B} \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{1-\sin^2 A - ( 1 - \sin^2 B) } { \sin^2 A - \sin^2 B} \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{1-\sin^2 A - 1 + \sin^2 B) } { \sin^2 A - \sin^2 B} \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{ \sin^2 B - \sin^2 A } { \sin^2 A - \sin^2 B} \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{ -( \sin^2 A - \sin^2 B) } { \sin^2 A - \sin^2 B} \times \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = - \frac{\sin A + \sin B}{\cos A - \cos B} \\ & = \frac{ \sin B + \sin A }{\cos B - \cos A }\\ & = \text{ RHS} \end{align}
1. (n) \( \frac{\sin^3 A + \cos^3 A }{\sin A + \cos A } + \frac{ \sin^3 A - \cos^3 A }{\sin A - \cos A } = 2 \)
Solution: \begin{align} \text{LHS } & = \frac{\sin^3 A + \cos^3 A }{\sin A + \cos A } + \frac{ \sin^3 A - \cos^3 A }{\sin A - \cos A } \\ & = \frac{(\sin A + \cos A)( \sin^2 A -\sin A \cos A + \cos^2 A )}{\sin A + \cos A } \\ & + \frac{(\sin A - \cos A ) ( \sin^2 A + \sin A \cos A + \cos^2 A )}{\sin A - \cos A } \\ & = \sin^2 - \sin A \cos A + \cos^2 A + \sin^2 A + \sin A \cos A + \cos^2 A \\ & = 2\sin^2 A + 2\cos^2 A \\ & = 2(\sin^2 A +\cos^2 A ) \\ & = 2(1) \\ & = 2\\ & = \text{ RHS} \end{align}
1. (o) \( \frac{\tan A - \cot A }{ \cos A . \sin^2 A - \sin A . \cos^2 A } = \sec A.\text{cosec} A ( \sec A + \text{cosec} A ) \)
Solution: \begin{align} \text{LHS } & = \frac{\tan A - \cot A }{ \cos A . \sin^2 A - \sin A . \cos^2 A } \\ & = \frac{\dfrac{\sin A }{\cos A } - \dfrac{\cos A }{\sin A} } { \cos A \sin A ( \sin A - \cos A )} \\ & = \frac{\dfrac{\sin^2 A - \cos^2 A }{ \cos A \sin A } } { \cos A \sin A ( \sin A - \cos A )} \\ & = \frac{\sin^2 A - \cos^2 A } { \cos^2 A \sin^2 A ( \sin A - \cos A )} \\ & = \frac{(\sin A - \cos A) (\sin A + \cos A ) } { \cos^2 A \sin^2 A ( \sin A - \cos A )} \\ & = \frac{\sin A + \cos A } { \cos^2 A \sin^2 A } \\ & = \frac{1}{\cos A \sin A} \times \frac{\sin A + \cos A}{ \cos A \sin A} \\ & = \sec A \text{cosec }A \times \left( \frac{\sin A}{\cos A \sin A} + \frac{\cos A}{\cos A \sin A} \right)\\ & = \sec A \text{cosec }A \times \left( \frac{1}{\cos A} + \frac{1}{\sin A } \right)\\ & = \sec A \text{cosec }A \left( \sec A + \text{cosec } A \right)\\ & = \text{ RHS} \end{align}
1. (p) \( \frac{1+\tan A . \tan B }{1-\tan A . \tan B } = \frac{\cos A. \cos B + \sin A . \sin B }{\cos A . \cos B - \sin A . \sin B } \)
Solution: \begin{align} \text{LHS } & = \frac{1+\tan A . \tan B }{1-\tan A . \tan B } \\ & = \frac{1 + \dfrac{\sin A \sin A}{ \cos A \cos B} }{1 - \dfrac{\sin A \sin B}{ \cos A \cos B} }\\ & = \frac{ \dfrac{\cos A \cos B + \sin A \sin B }{\cos A \cos B} }{\dfrac{\cos A \cos B - \sin A \sin B }{\cos A \cos B} }\\ & = \frac{\cos A. \cos B + \sin A . \sin B }{\cos A . \cos B - \sin A . \sin B }\\ & = \text{ RHS} \end{align}
1. (q) \( \frac{\cos A }{ \sin A + \cos B } + \frac{\cos B }{\sin B - \cos A} = \frac{ \cos A }{ \sin A - \cos B } + \frac{ \cos B }{ \sin B + \cos A } \)
Solution: \begin{align} & \frac{\cos A }{ \sin A + \cos B } + \frac{\cos B }{\sin B - \cos A} = \frac{ \cos A }{ \sin A - \cos B } + \frac{ \cos B }{ \sin B + \cos A } \\ & \text{By Transposition, We get } \\ & \frac{\cos A }{ \sin A + \cos B } - \frac{\cos A}{\sin A - \cos B} = \frac{ \cos B }{ \sin B + \cos A } - \frac{\cos B}{\sin B - \cos A}\\ \text { New LHS }& = \frac{\cos A }{ \sin A + \cos B } - \frac{\cos A}{\sin A - \cos B}\\ & = \frac{\cos A ( \sin A - \cos B ) - \cos A ( \sin A + \cos B) }{(\sin A + \cos B) ( \sin A - \cos B) } \\ & =\frac{\sin A \cos A - \cos A \cos B - \sin A \cos A - \cos A \cos B }{\sin^2 A - \cos^2 B } \\ & = \frac{-2\cos A \cos B} { \sin^2 A - \cos^2 B} \\ & = \frac{2\cos A \cos B}{ \cos^2 B - \sin^2 A} \\ \text{New RHS} & = \frac{ \cos B }{ \sin B + \cos A } - \frac{\cos B}{\sin B - \cos A}\\ & = \frac{ \cos B( \sin B - \cos A ) - \cos B(\sin B + \cos A )}{ (\sin B + \cos A ) ( \sin B - \cos A ) } \\ & = \frac{\sin B \cos B - \cos A \cos B - \sin B \cos B + \cos A \cos B }{ \sin^2 B - \cos^2 A } \\ & = \frac{-2\cos A \cos B}{ \sin^2 B - \cos^2 A } \\ & = \frac{2 \cos A \cos B}{ \cos^2 A - \sin^2 B } \\ & = \frac{2\cos A \cos B }{ 1-\sin^2 A - ( 1 - \cos^2 B ) } \\ & = \frac{2\cos A \cos B}{ 1 -\sin^2 A - 1 + \cos^2 B } \\ & = \frac{ 2\cos A \cos B}{ \cos^2 B - \sin^2 A } \\ &\text{Hence, New LHS = New RHS } \end{align}
1. (r) \( \frac{1}{\sec A - \tan A } - \frac{1}{\cos A } = \frac{1}{ \cos A } - \frac{1}{\sec A + \tan A } \)
Solution: \begin{align} & \frac{1}{\sec A - \tan A } - \frac{1}{\cos A } = \frac{1}{ \cos A } - \frac{1}{\sec A + \tan A } \\ & \text{By Transposition, We get } \\ & \frac{1}{\sec A - \tan A } + \frac{1}{ \sec A + \tan A } = \frac{1}{\cos A} + \frac{1}{\cos A } \\ \text { New LHS }& = \frac{1}{\sec A - \tan A } + \frac{1}{ \sec A + \tan A } \\ & = \frac{\sec A + \tan A + \sec A - \tan A }{ (\sec A - \tan A )( \sec A + \tan A )} \\ & = \frac{2\sec A }{ \sec^2 A -\tan^2 A } \\ & = \frac{2 \sec A }{ 1 } \\ & = 2\sec A \\ \text{New RHS} & = \frac{1}{\cos A} + \frac{1}{\cos A } \\ & = \sec A + \sec A \\ & = 2\sec A \\ &\text{Hence, New LHS = New RHS } \end{align}
2. (a) \( \frac{\tan A - \sec A + 1}{\tan A + \sec A - 1 } = \sec A - \tan A \)
Solution: \begin{align} \text{LHS } & = \frac{\tan A - \sec A + 1}{\tan A + \sec A - 1 } \\ & = \frac{1 - \sec A +\tan A}{\sec A + \tan A - 1 } \\ & = \frac{\sec^2 A - \tan^2 A - ( \sec A - \tan A ) }{ \sec A + \tan A - 1 } \\ & = \frac{(\sec A - \tan A ) ( \sec A + \tan A ) - 1 ( \sec A - \tan A ) }{ \sec A + \tan A - 1 } \\ & = \frac{ ( \sec A - \tan A ) ( \sec A + \tan A - 1 )}{ \sec A + \tan A - 1 } \\ & = \sec A - \tan A \\ & = \text{RHS } \end{align}
2. (b) \( \frac{\sec A + \cos A - 1 }{ 1 - \sec A + \tan A} = \sec A + \tan A \)
Solution: \begin{align} \text{LHS } & = \frac{\sec A + \cos A - 1 }{ 1 - \sec A + \tan A} \\ & = \frac{ \sec A +\tan A - 1}{1 - \sec A + \tan A } \\ & = \frac{ ( \sec A + \tan A ) - (\sec^2 A - \tan^2 A )}{ 1 - \sec A + \tan A} \\ & = \frac{ ( \sec A + \tan A ) - (\sec A - \tan A ) ( \sec A + \tan A ) }{1 - \sec A + \tan A } \\ & = \frac{ ( \sec A + \tan A ) ( 1 - \sec A + \tan A )}{ 1 - \sec A + \tan A } \\ & = \sec A + \tan A \\ & = \text{RHS } \end{align}
2. (c) \( \frac{\sin A + \cos A - 1 }{\sin A - \cos A + 1 } = \frac{ 1- \sin A }{\cos A } \)
Solution: \begin{align} \text{LHS } & = \frac{\sin A + \cos A - 1 }{\sin A - \cos A + 1 } \\ & = \frac{\dfrac{\sin A } {\cos A} + \dfrac{\cos A }{\cos A} - \dfrac{ 1}{ \cos A } }{\dfrac{\sin A}{\cos A} - \dfrac{\cos A}{\cos A} + \dfrac{1}{\cos A} } \\ & = \frac{\tan A + 1 - \sec A }{\tan A - 1 + \sec A } \\ & = \frac{1 - \sec A + \tan A }{\sec A + \tan A - 1 } \\ & = \frac{ \sec^2 A - \tan^2 A - ( \sec A - \tan A) }{\sec A + \tan A - 1 } \\ & = \frac{ ( \sec A - \tan A) (\sec A + \tan A )- ( \sec A - \tan A)}{\sec A + \tan A - 1 } \\ & = \frac{(\sec A - \tan A) ( \sec A + \tan A - 1 )}{\sec A + \tan A - 1 }\\ & = \sec A - \tan A \\ & = \frac{1}{\cos A } - \frac{\sin A }{\cos A } \\ & = \frac{ 1 - \sin A }{ \cos A } \\ & = \text{RHS } \end{align}
2. (d) \( \frac{1-\sin A - \cos A }{\cos A - \sin A - 1} = \frac{1-\sin A }{\cos A } \)
Solution: \begin{align} \text{LHS } & = \frac{1-\sin A - \cos A }{\cos A - \sin A - 1} \\ & =\frac{\dfrac{1}{\cos A} - \dfrac{\sin A}{ \cos A} - \dfrac{\cos A }{ \cos A} }{ \dfrac{\cos A }{ \cos A } - \dfrac{\sin A }{ \cos A} - \dfrac{1}{ \cos A} } \\ & = \frac{ \sec A - \tan A - 1 }{ 1 - \tan A - \sec A } \\ & = \frac{ \sec A - \tan A - ( \sec^2 A - \tan^2 A ) }{ 1 - \tan A - \sec A } \\ & = \frac{(\sec A - \tan A ) - ( \sec A - \tan A ) (\sec A + \tan A ) }{ 1 - \tan A - \sec A } \\ & = \frac{ ( \sec A -\tan A ) ( 1 - \sec A - \tan A ) }{ 1 - \tan A - \sec A } \\ & = \sec A - \tan A \\ & = \text{RHS } \end{align}
2. (e) \( \frac{1-\text{cosec } A + \cot A }{ \text{cosec } A + \cot A - 1 } = \text{cosec } A - \cot A \)
Solution: \begin{align} \text{LHS } & = \frac{1-\text{cosec } A + \cot A }{ \text{cosec } A + \cot A - 1 } \\ & = \frac{ (\text{cosec}^2 A - \cot^2 A ) - (\text{cosec }A - \cot A ) }{ \text{cosec } A + \cot A - 1 } \\ & = \frac{ ( \text{cosec}A - \cot A ) ( \text{cosec} A + \cot A ) - (\text{cosec }A - \cot A ) }{ \text{cosec } A + \cot A - 1 } \\ & = \frac{(\text{cosec} A - \cot A ) ( \text{cosec} A + \cot A - 1 ) }{ \text{cosec} A + \cot A - 1 } \\ & = \text{cosec} A - \cot A \\ & = \text{RHS } \end{align}
2. (f) \( \frac{\cot A + \text{cosec }A - 1 }{ \cot A - \text{cosec } A + 1 } = \frac{1+\cos A }{ \sin A } \)
Solution: \begin{align} \text{LHS } & = \frac{\cot A + \text{cosec }A - 1 }{ \cot A - \text{cosec } A + 1 } \\ & = \frac{ \text{cosec }A + \cot A - 1 }{ \cot A - \text{cosec } A + 1 }\\ & = \frac{ \text{cosec}A + \cot A - (\text{cosec}^2 A - \cot^2 A ) }{ \cot A - \text{cosec } A + 1 }\\ & = \frac{ (\text{cosec}A + \cot A) - (\text{cosec} A - \cot A )(\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 \\ & = \frac{1}{\sin A } + \frac{ \cos A }{ \sin A } \\ & = \frac{1+\cos A } {\sin A } \\ & = \text{RHS } \end{align}
2. (g) \( \frac{1+\cos \theta - \sin \theta } { \cos \theta + \sin \theta - 1 } = \frac{1 + \cos \theta }{ \sin \theta } \)
Solution: \begin{align} \text{LHS } & = \frac{1+\cos \theta - \sin \theta } { \cos \theta + \sin \theta - 1 } \\ & = \frac{\dfrac{1}{\sin \theta } + \dfrac{ \cos \theta }{ \sin \theta } - \dfrac{\sin \theta}{ \sin \theta }}{ \dfrac{\cos \theta }{ \sin \theta } + \dfrac{ \sin \theta }{ \sin \theta } - \dfrac{1}{\sin \theta } } \\ & = \frac{\text{cosec} \theta + \cot \theta - 1}{\cot \theta + 1 - \text{cosec} \theta } \\ & = \frac{\text{cosec} \theta + \cot \theta - ( \text{cosec}^2 \theta - \cot^2 \theta ) }{\cot \theta + 1 - \text{cosec} \theta } \\ & = \frac{\text{cosec} \theta + \cot \theta - (\text{cosec}\theta - \cot \theta ) ( \text{cosec} \theta + \cot \theta )}{\cot \theta + 1 - \text{cosec} \theta } \\ & = \frac{(\text{cosec} \theta + \cot \theta ) ( 1 - \text{cosec}\theta + \cot \theta )}{( 1 - \text{cosec}\theta + \cot \theta )} \\ & = \text{cosec} \theta + \cot \theta \\ & = \frac{1}{\sin \theta } + \frac{\cos \theta }{ \sin \theta } \\ & = \frac{ 1+\cos \theta }{ \sin \theta }\\ & = \text{RHS } \end{align}
2. (h) \( \frac{ \cos A + \sin A - 1 }{ \cos A - \sin A + 1 } = \frac{1-\cos A }{ \sin A } \)
Solution: \begin{align} \text{LHS } & = \frac{ \cos A + \sin A - 1 }{ \cos A - \sin A + 1 } \\ & = \frac{ \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\sin A } - \dfrac{1}{\sin A} }{ \dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\sin A} + \dfrac{1}{\sin A} } \\ & = \frac{ \cot A + 1 - \text{cosec} A }{ \cot A - 1 + \text{cosec} A }\\ & = \frac{ 1 - \text{cosec} A + \cot A }{ \cot A - 1 + \text{cosec} A }\\ & = \frac{ \text{cosec}^2 A - \cot^2 A - \text{cosec} A + \cot A }{ \cot A - 1 + \text{cosec} A }\\ & = \frac{ (\text{cosec} A - \cot A ) (\text{cosec} A +\cot A ) - (\text{cosec} A - \cot A) }{ \cot A - 1 + \text{cosec} A }\\ & = \frac{ (\text{cosec} A - \cot A ) (\text{cosec} A +\cot A - 1) }{ \text{cosec} A + \cot A - 1 }\\ & = \text{cosec} A - \cot A \\ & = \frac{1}{\sin A} - \frac{\cos A }{ \sin A} \\ & = \frac{1-\cos A }{ \sin A } \\ & = \text{RHS } \end{align}
2. (i) \( (3 - 4 \sin^2 A ) ( 2 - \tan^2 A ) = ( 2 - 3 \sin^2 A ) ( 3 - \tan^2 A ) \)
Solution: \begin{align} \text{LHS} & = (3 - 4 \sin^2 A ) ( 2 - \tan^2 A ) \\ & = (3 - 4 \sin^2 A ) \left( 2 - \frac{\sin^2 A }{ \cos^2 A } \right)\\ & = (3 - 4 \sin^2 A ) \left( \frac{2\cos^2 A - \sin^2 A }{ \cos^2 A }\right) \\ & = \frac{3-4\sin^2 A}{\cos^2 A } \left(2\cos^2 A - \sin^2 A \right)\\ & = \left(\frac{3}{\cos^2 A } - \frac{4\sin^2 A }{ \cos^2 A }\right) \left\{ 2(1-\sin^2 A) - \sin^2 A \right\} \\ & = ( 3\sec^2 A - 4\tan^2 A) ( 2-2\sin^2 A - \sin^2 A ) \\ & = \left\{ 3(1+\tan^2 A ) - 4 \tan^2 A\right\} ( 2 - 3 \sin^2 A )\\ & = ( 3+3\tan^2 A - 4 \tan^2 A )( 2 - 3 \sin^2 A )\\ & = ( 3 - \tan^2 A ) ( 2 - 3 \sin^2 A )\\ & = ( 2 - 3 \sin^2 A ) ( 3 - \tan^2 A ) \\ & = \text{ RHS} \end{align}
2. (j) \( ( 1 - 2 \cos^2 A ) ( 3 - 2\cot^2 A ) = ( 1 - \cot^2 A )( 3 - 5\cos^2 A ) \)
Solution: \begin{align} \text{LHS} & = ( 1 - 2 \cos^2 A ) ( 3 - 2\cot^2 A ) \\ & = ( 1 - 2 \cos^2 A ) \left( 3 - \frac{2\cos^2 A}{\sin^2 A} \right) \\ & = ( 1 - 2 \cos^2 A ) \left( \frac{3\sin^2 A - 2\cos^2 A}{\sin^2 A} \right) \\ & = \left( \frac{1-2\cos^2 A}{\sin^2 A} \right) ( 3\sin^2 A - 2\cos^2 A ) \\ & = \left( \frac{1}{\sin^2 A} - \frac{2\cos^2 A }{ \sin^2 A } \right) \left\{ 3(1-\cos^2 A) - 2\cos^2 A \right\} \\ & = \left( \text{cosec}^2 A - 2\cot^2 A \right)( 3 - 3\cos^2 A - 2 \cos^2 A ) \\ & = \left( 1+\cot^2 A - 2\cot^2 A \right)( 3 - 5\cos^2 A ) \\ & = \left( 1 - \cot^2 A \right)( 3 - 5\cos^2 A ) \\ & = \text{ RHS} \end{align}
2. (k) \( (5 + 2 \sin^2 A )( 3 \tan^2 A - 1 ) = ( 7\sec^2 A - 2 ) (3 - 4 \cos^2 A ) \)
Solution: \begin{align} \text{LHS} & =(5 + 2 \sin^2 A )( 3 \tan^2 A - 1 )\\ & = (5 + 2 \sin^2 A ) \left( \frac{3\sin^2 A }{\cos^2 A} - 1 \right)\\ & = (5 + 2 \sin^2 A )\left( \frac{3 \sin^2 A - \cos^2 A}{ \cos^2 A } \right)\\ & = \left(\frac{5+2\sin^2 A }{ \cos^2 A }\right) (3 \sin^2 A - \cos^2 A)\\ & = \left( \frac{5}{\cos^2 A} + \frac{2\sin^2 A}{\cos^2 A } \right) \left\{ 3(1-\cos^2 A) - \cos^2 A \right \} \\ & = ( 5\sec^2 A + 2\tan^2 A) ( 3 - 3\cos^2 A - \cos^2 A ) \\ & = \left\{ 5\sec^2 A + 2(\sec^2 A - 1 ) \right\} ( 3 - 4\cos^2 A )\\ & = ( 5\sec^2 A + 2 \sec^2 A - 2 ) ( 3-4\cos^2 A ) \\ & = ( 7\sec^2 A - 2 ) (3 - 4 \cos^2 A ) \\ & = \text{ RHS} \end{align}
2. (l) \( (2 - \cot^2 A ) ( 1 - 3 \cos^2 A ) = ( 3 \sin^2 A - 1 ) ( 1 - 2 \cot^2 A ) \)
Solution: \begin{align} \text{LHS} & = (2 - \cot^2 A ) ( 1 -3 \cos^2 A ) \\ & = \left( 2 - \frac{\cos^2 A }{ \sin^2 A } \right) ( 1 - 3 \cos^2 A ) \\ & = \left( \frac{2\sin^2 A - \cos^2 A }{ \sin^2 A } \right) ( 1 - 3 \cos^2 A ) \\ & = ( 2 \sin^2 A - \cos^2 A ) \left( \frac{1-3\cos^2 A }{\sin^2 A } \right) \\ & = \left\{ 2\sin^2 A - ( 1 - \sin^2 A ) \right\} \left( \frac{1}{\sin^2 A } - \frac{3\cos^2 A } { \sin^2 A } \right) \\ & = ( 2\sin^2 A - 1 + \sin^2 A )( \text{cosec}^2 A - 3\cot^2 A ) \\ & = (3\sin^2 A - 1 ) ( 1 + \cot^2 A - 3 \cot^2 A )\\ & = ( 3\sin^2 A - 1 ) ( 1 - 2\cot^2 A ) \\ \end{align}
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