CDC Opt. Maths
Exercise 5.4 [ Page - 191 ] [ English medium]
1. Answer the following:
(a) Define trigonometric identity.
Solution: Any identity containing the trigonometric ratios like sin, cos and tan are called trigonometric identities.
(b) Write the relation between \( \sin \theta \text{ and } \cos \theta \).
Solution: \( \sin \theta = \sqrt{1-\cos^2 \theta } \)
(c) Write \( \sec \theta \) in terms of \( \tan \theta \).
Solution: \( \sec \theta = \sqrt{1+\tan^2 \theta } \)
(d) What is the product of \( ( \text{cosec } \theta + \cot \theta ) \text{ and } (\text{ cosec } \theta - \cot \theta ) \)?
Solution: \begin{align} \ & \ \ ( \text{cosec } \theta + \cot \theta ) (\text{ cosec } \theta - \cot \theta ) \\ \ & = \text{cosec}^2 \theta - \cot^2 \theta \\ \ & = 1 \\ \end{align}
2. (a) Multiply: \( ( \sin A +\sin B ) ( \sin A -\sin B ) \)
Solution: \begin{align} \ & ( \sin A +\sin B ) ( \sin A -\sin B ) \\ \ & = \sin^2 A - \sin^2 B \\ \end{align}
2. (b) Multiply: \( (1-cos \alpha ) (1+\cos \alpha ) \)
Solution: \begin{align} \ & (1- \cos \alpha ) (1+\cos \alpha ) \\ \ & = 1^2 - \cos^2 \alpha \\ \ & = 1 -\cos^2 \alpha \\ \end{align}
2. (c) Multiply: \( ( 1+\cos x ) (1-cos x ) \)
Solution: \begin{align} \ & ( 1+\cos x ) (1-\cos x ) \\ \ & = 1^2 -\cos^2 x \\ \ & = 1 - \cos^2 x \\ \end{align}
2. (d) Multiply: \( (1+\tan^2 A ) (1-\tan^2 A ) \)
Solution: \begin{align} \ & (1+\tan^2 A ) (1-\tan^2 A ) \\ \ & = 1^2 - (\tan^2 A )^2 \\ \ & = 1 - \tan^4 A \end{align}
2. (e) Multiply: \( (1+\sin \theta ) (1-\sin \theta ) (1+\sin^2 \theta ) \)
Solution: \begin{align} \ & (1+\sin \theta ) (1-\sin \theta ) (1+\sin^2 \theta ) \\ \ & = (1^2 - \sin^2 \theta ) (1+\sin^2 \theta) \\ \ & = (1-\sin^2 \theta )(1+\sin^2 \theta )\\ \ & = 1^2 -(\sin^2 \theta )^2 \\ \ & = 1 - \sin^4 \theta \end{align}
2. (f) Multiply: \( (1+\tan \alpha ) (1-\tan \alpha ) (1+\tan^2 \alpha ) \)
Solution: \begin{align} \ & (1+\tan \theta ) (1-\tan \theta ) (1+\tan^2 \theta ) \\ \ & = (1^2 - \tan^2 \theta ) (1+\tan^2 \theta) \\ \ & = (1-\tan^2 \theta )(1+\tan^2 \theta )\\ \ & = 1^2 -(\tan^2 \theta )^2 \\ \ & = 1 - \tan^4 \theta \end{align}
3. (a) Factorize: \( \tan^2 A -\sin^2 A \)
Solution: \begin{align} \ & \tan^2 A -\sin^2 A \\ \ & =(\tan A -\sin A)(\tan A + \sin A) \\ \end{align}
3. (b) Factorize: \( \cos^2 A -\sec^2 A \)
Solution: \begin{align} \ & \cos^2 A -\sec^2 A \\ \ & = ( \cos A -\sec A )( \cos A + \sec A ) \\ \end{align}
3. (c) Factorize: \( \sin^2 x + \cos^2 x. \sin^2 x \)
Solution: \begin{align} \ & \sin^2 x + \cos^2 x. \sin^2 x \\ \ & = \sin^2 x ( 1+\cos^2 x) \\ \end{align}
3. (d) Factorize: \( \sin^3 \alpha - \cos^3 \alpha \)
Solution: \begin{align} \ & \sin^3 \alpha - \cos^3 \alpha \\ \ & = (\sin \alpha - \cos \alpha ) (\sin^2 \alpha +\sin \alpha \cos \alpha + \cos^2 \alpha) \\ \ & = (\sin \alpha - \cos \alpha ) (\sin^2 \alpha +\cos^2 \alpha +\sin \alpha \cos \alpha ) \\ \ & = (\sin \alpha - \cos \alpha ) (1 +\sin \alpha \cos \alpha ) \\ \end{align}
3. (e) Factorize: \( \sec^4 \theta - \cos^4 \theta \)
Solution: \begin{align} \ & \sec^4 \theta - \cos^4 \theta \\ \ & = (\sec^2 \theta)^2 - (\cos^2 \theta )^2 \\ \ & = (\sec^2 \theta -\cos^2 \theta )( \sec^2 \theta + \cos^2 \theta )\\ \ & = (\sec \theta -\cos \theta )(\sec \theta + \cos \theta ) (\sec^2 \theta + \cos^2 \theta ) \end{align}
3. (f) Factorize: \( \sin^2 x + 5 \sin x + 6 \)
Solution: \begin{align} \ & \sin^2 x + 5 \sin x + 6 \\ \ & = \sin^2 x + (3+2) \sin x + 6 \\ \ & = \sin^2 x + 3\sin x + 2\sin x + 6 \\ \ & = \sin x (\sin x +3 ) +2 ( \sin x +3 ) \\ \ & = (\sin x +3 )( \sin x +2 ) \\ \end{align}
4.(a) Prove that: \( \tan A \times \cos A = \sin A \)
Solution: \begin{align} \text{LHS} & = \tan A \times \cos A \\ \ & = \frac{\sin A}{\cos A} \times \cos A \\ \ & = \sin A\\ \ & = \text{RHS}\\ \end{align}
4.(b) Prove that: \( \cos \theta \times \text{ cosec } \theta = \cot \theta \)
Solution: \begin{align} \text{LHS} & = \cos \theta \times \text{ cosec } \theta = \cot \theta \\ \ & = \cos \theta \times \frac{1}{\sin \theta } \\ \ & = \frac{\cos \theta }{\sin \theta } \\ \ & = \cot \theta \\ \ & = \text{RHS}\\ \end{align}
4.(c) Prove that: \( \sec \alpha \times \sin \alpha \times \cot \alpha = 1 \)
Solution: \begin{align} \text{LHS} & = \sec \alpha \times \sin \alpha \times \cot \alpha \\ \ & = \frac{1}{\cos \alpha} \times \sin \alpha \times \frac{\cos \alpha} {\sin \alpha} \\ \ & = 1 \\ \ & = \text{RHS}\\ \end{align}
4.(d) Prove that: \( \cot \beta \times \sin \beta = \cos \beta \)
Solution: \begin{align} \text{LHS} & = \cot \beta \times \sin \beta \\ \ & = \frac{\cos A}{\sin A} \times \sin A \\ \ & = \cos A\\ \ & = \text{RHS}\\ \end{align}
4.(e) Prove that: \( \frac{\sin A \times \text{ cosec } A }{\tan A } = \cot A \)
Solution: \begin{align} \text{LHS} & = \frac{\sin A \times \text{ cosec } A }{\tan A } \\ \ & = \frac{1}{\tan A}\\ \ & = \cot A \\ \ & = \text{RHS}\\ \end{align}
4.(f) Prove that: \( \frac{\cot A \times \tan A } {\sin A \times \cos A } = \text{cosec } A \sec A \) Solution: \begin{align} \text{LHS} & = \frac{\cot A \times \tan A } {\sin A \times \cos A } \\ \ & = \frac{1 } {\sin A \times \cos A } \\ \ & = \text{cosec } A \sec A\\ \ & = \text{RHS}\\ \end{align}
5. (a) Prove that: \( \cos^2 A - \cos^2 A \times \sin^2 A = \cos^4 A \)
Solution: \begin{align} \text{LHS} & = \cos^2 A - \cos^2 A \times \sin^2 A \\ \ & = \cos^2 A (1-\sin^2 A ) \\ \ & = \cos^2 A \times \cos^2 A\\ \ & = \cos^4 A \\ \ & = \text{RHS}\\ \end{align}
5. (b) Prove that: \( (1-\cos^2 \theta ) (1+\tan^2 \theta ) = \tan^2 \theta \)
Solution: \begin{align} \text{LHS} & = (1-\cos^2 \theta ) (1+\tan^2 \theta ) \\ \ & =\sin^2 \theta \times \sec^2 \theta \\ \ & = \sin^2 \theta \times \frac{1}{\cos^2 \theta } \\ \ & = \frac{\sin^2 \theta}{\cos^2 \theta } \\ \ & = \tan^2 \theta \\ \ & = \text{RHS}\\ \end{align}
5. (c) Prove that: \( \sin^2 A \times \cos^2 A + \sin^4 A = \sin^2 A \)
Solution: \begin{align} \text{LHS} & = \sin^2 A \times \cos^2 A + \sin^4 A \\ \ & = \sin^2 A ( \cos^2 A + \sin^2 A ) \\ \ & = \sin^2 A (1) \\ \ & = \sin^2 A \\ \ & = \text{RHS}\\ \end{align}
5. (d) Prove that: \( (\cot^2 \alpha +1) \times \tan^2 \alpha = \sec^2 \alpha \)
Solution: \begin{align} \text{LHS} & = (\cot^2 \alpha +1) \times \tan^2 \alpha \\ \ & = \text{cosec } \alpha \times \tan^2 \alpha \\ \ & = \frac{1}{\sin \alpha } \times \frac{\sin \alpha}{\cos \alpha }\\ \ & = \frac{1}{\cos \alpha } \\ \ & = \sec \alpha \\ \ & = \text{RHS}\\ \end{align}
5. (e) Prove that: \( (1+\sin A )^2 - (1-\sin A )^2 =4\sin A \)
Solution: \begin{align} \text{LHS} & =(1+\sin A )^2 - (1-\sin A )^2 \\ \ & = 1^2 +2\times 1\times \sin A + \sin^2 A - (1^2 -2\times 1\times \sin A + \sin^2 A ) \\ \ & = 1+2\sin A + \sin^2 A - 1 +2\sin A - \sin^2 A \\ \ & = 4\sin A \\ \ & = \text{RHS}\\ \end{align}
5. (f) Prove that: \( (1+\tan \alpha )^2 + (1-\tan \alpha )^2 = 2\sec^2 \alpha \)
Solution: \begin{align} \text{LHS} & = (1+\tan \alpha )^2 + (1-\tan \alpha )^2 \\ \ & = 1^2 +2\times 1 \times \tan \alpha +\tan^2 \alpha + 1^2 - 2\times 1\times \tan \alpha + \tan^2 \alpha \\ \ & = 1+\tan^2 \alpha + 1+ \tan^2 \alpha\\ \ & = \sec^2 \alpha + \sec^2 \alpha \\ \ & = 2\sec^2 \alpha \\ \ & = \text{RHS}\\ \end{align}
5. (g) Prove that: \( \frac{\text{ cosec}^2\theta-1}{\text{cosec}^2 \theta } =\cos^2 \theta \)
Solution: \begin{align} \text{LHS} & =\frac{\text{ cosec}^2\theta-1}{\text{cosec}^2 \theta } \\ \ & = \frac{\text{ cosec}^2\theta } {\text{ cosec}^2\theta } - \frac{1}{\text{ cosec}^2\theta} \\ \ & = 1 - \sin^2 \theta \\ \ & =\cos^2 \theta \\ \ & = \text{RHS}\\ \end{align}
5. (h) Prove that: \( \tan^2 \alpha - \sin^2 \alpha = \tan^2 \alpha \times \sin^2 \alpha \)
Solution: \begin{align} \text{LHS} & =\tan^2 \alpha - \sin^2 \alpha \\ \ & = \frac{\sin^2 \alpha } {\cos^2 \alpha } - \sin^2 \alpha \\ \ & = \sin^2 \alpha \times \sec^2 \alpha - \sin^2 \alpha \\ \ & = \sin^2 \alpha ( \sec^2 \alpha - 1) \\ \ & = \sin^2 \alpha \tan^2 \alpha\\ \ & = \tan^2 \alpha \sin^2 \alpha \\ \ & = \text{RHS}\\ \end{align}
5. (i) Prove that: \( \cos \theta \sqrt{1+\cot^2 \theta} = \sqrt{ \text{cosec}^2 \theta -1 } \)
Solution: \begin{align} \text{LHS} & = \cos \theta \sqrt{1+\cot^2 \theta} \\ \ & = \cos \theta \sqrt{\text{cosec}^2 \theta} \\ \ & = \cos \theta \text{cosec} \theta\\ \ & = \cos \theta \times \frac{1}{\sin \theta }\\ \ & = \frac{\cos \theta }{ \sin \theta } \\ \ & = \cot \theta \\ \ & = \text{RHS}\\ \end{align}
5. (j) Prove that: \( \sqrt{ 1+\tan^2 \alpha} \times \sqrt{1-\cos^2 \alpha } = \tan \alpha \)
Solution: \begin{align} \text{LHS} & = \sqrt{ 1+\tan^2 \alpha} \times \sqrt{1-\cos^2 \alpha } \\ \ & = \sqrt{\sec^2 \alpha } \times \sqrt{\sin^2 \alpha} \\ \ & = \sec \alpha \times \sin \alpha \\ \ & = \frac{1}{\cos \alpha} \times \sin \alpha \\ \ & = \frac{\sin \alpha } {\cos \alpha} \\ \ & = \tan \alpha \\ \ & = \text{RHS}\\ \end{align}
6. (a) Prove that: \( \frac{1 - \sin^4 A }{\cos^4 A } = 1 + 2\tan^2 A\)
Solution: \begin{align} \text{LHS} & = \frac{1 - \sin^4 A }{\cos^4 A }\\ \ & = \frac{ 1^2 - (\sin^2 A )^2 }{\cos^4 A} \\ \ & = \frac{ (1-\sin^2 A)(1+\sin^2 A) } {\cos^4 A} \\ \ & = \frac{ \cos^2 A (1+\sin^2 A)}{\cos^4 A} \\ \ & = \frac{1+\sin^2 A }{\cos^2 A}\\ \ & = \frac{1}{\cos^2 A} + \frac{\sin^2 A}{\cos^2 A} \\ \ & = \sec^2 A + \tan^2 A \\ \ & = 1 + \tan^2 A + \tan^2 A \\ \ & = 1 + 2\tan^2 A \\ \ & = \text{RHS}\\ \end{align}
6. (b) Prove that: \( \frac{1-\tan^4 A }{\sec^4 A } = 1 - 2\sin^2 A \)
Solution: \begin{align} \text{LHS} & = \frac{1-\tan^4 A }{\sec^4 A }\\ \ & = \frac{1^2 - (\tan^2 A )^2 } { \sec^4 A } \\ \ & = \frac{ (1-\tan^2 A)(1+\tan^2 A)}{\sec^4 A} \\ \ & = \frac{ (1 -\tan^2 A) \sec^2 A }{\sec^4 A} \\ \ & = \frac{ 1- \tan^2 A}{\sec^2 A}\\ \ & = \frac{1}{\sec^2 A } - \frac{\tan^2 A}{\sec^2 A}\\ \ & = \cos^2 A - \dfrac{\dfrac{\sin^2 A}{\cos^2 A}}{\dfrac{1}{\cos^2 A }}\\ \ & = \cos^2 A - \sin^2 A \\ \ & = 1-\sin^2 A - \sin^2 A \\ \ & = 1 - 2\sin^2 A \\ \ & = \text{RHS}\\ \end{align}
6. (c) Prove that: \( \cos^4 \theta - \sin^4 \theta = \cos^2 \theta - \sin^2 \theta \)
Solution: \begin{align} \text{LHS} & = \cos^4 \theta - \sin^4 \theta \\ \ & = (\cos^2 \theta )^2 - (\sin^2 \theta )^2 \\ \ & = (\cos^2 \theta - \sin^2 \theta )( \cos^2 \theta + \sin^2 \theta ) \\ \ & = (\cos^2 \theta -\sin^2 \theta ) (1)\\ \ & = \cos^2 \theta - \sin^2 \theta \\ \ & = \text{RHS}\\ \end{align}
6. (d) Prove that: \( \frac{\sin^3 \alpha+\cos^3 \alpha }{\sin \alpha + \cos \alpha } = 1- 2\sin \alpha . \cos \alpha \)
Solution: \begin{align} \text{LHS} & = \frac{\sin^3 \alpha+\cos^3 \alpha }{\sin \alpha + \cos \alpha } \\ \ & = \frac{(\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha )}{\sin \alpha + \cos \alpha } \\ \ & = \sin^2 \alpha + \cos^2 \alpha -\sin \alpha \cos \alpha \\ \ & = 1- 2\sin \alpha . \cos \alpha \\ \ & = \text{RHS}\\ \end{align}
6. (e) Prove that: \( \frac{\sin^3 A - \cos^3 A}{1-\sin A \times \cos A } = \sin A - \cos A \)
Solution: \begin{align} \text{LHS} & = \frac{\sin^3 A - \cos^3 A}{1-\sin A \times \cos A }\\ \ & = \frac{(\sin A -\cos A)(\sin^2 A + \sin A \cos A + \cos^2 A )}{1-\sin A \cos A } \\ \ & = \frac{(\sin A - \cos A ) ( \sin^2 + \cos^2 A - \sin A \cos A) }{1- \sin A \cos A }\\ \ & = \frac{(\sin A - \cos A ) ( 1 - \sin A \cos A) }{1- \sin A \cos A }\\ \ & = \sin A - \cos A \\ \ & = \text{RHS}\\ \end{align}
6. (f) Prove that: \( \frac{\tan A - 1} { \tan A + 1} = \frac{ 2\sin^2 A - 1}{1+ 2\sin A \times \cos A }\) Solution: \begin{align} \text{LHS} & = \frac{\tan A - 1} { \tan A + 1} \\ \ & = \dfrac{\dfrac{\sin A }{ \cos A} -1 }{\dfrac{\sin A }{ \cos A} + 1 } \\ \ & = \dfrac{\dfrac{\sin A - \cos A}{\cos A}}{\dfrac{\sin A + \cos A}{\cos A}}\\ \ & = \frac{\sin A - \cos A}{\sin A + \cos A }\\ \ & = \frac{\sin A - \cos A}{\sin A + \cos A }\times \frac{\sin A + \cos A}{\sin A + \cos A }\\ \ & = \frac{\sin^2 A - \cos^2 A}{( \sin A + \cos A )^2 }\\ \ & = \frac{\sin^2 A - (1-\sin^2 A)}{ \sin^2 A +2\sin A \cos A + \cos^2 A }\\ \ & = \frac{\sin^2 A - 1+ \sin^2 A)}{ \sin^2 A +\cos^2 A+2\sin A \cos A }\\ \ & = \frac{ 2\sin^2 A - 1}{1+ 2\sin A \times \cos A } \\ \ & = \text{RHS}\\ \end{align}
6. (g) Prove that: \( \frac{1}{\sec \theta + \tan \theta } = \frac{ 1-\sin \theta}{\cos \theta } \)
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 \\ \ & = \frac{1}{\cos \theta } - \frac{\sin \theta }{ \cos \theta } \\ \ & = \frac{ 1-\sin \theta }{\cos \theta } \\ \ & = \text{RHS}\\ \end{align}
6. (h ) Prove that: \( \frac{1}{\tan \theta + \cot \theta } = \sin \theta \times \cos \theta \)
Solution: \begin{align} \text{LHS} & = \frac{1}{\tan \theta + \cot \theta } \\ \ & = \dfrac{1}{\dfrac{\sin \theta}{\cos \theta } + \dfrac{\cos \theta }{\sin \theta } } \\ \ & = \dfrac{1}{\dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta }}\\ \ & = \frac{\sin \theta \cos \theta }{\sin^2 \theta + \cos^2 \theta } \\ \ & = \frac{\sin \theta \cos \theta }{1} \\ \ & = \sin \theta \times \cos \theta \\ \ & = \text{RHS}\\ \end{align}
6. (i) Prove that: \( \frac{1}{\sec \theta - \tan \theta} = \frac{1+\sin \theta}{\cos \theta } = \sec \theta + \tan \theta \)
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} Also, \begin{align} \text{Mid term } & = \frac{1+\sin \theta}{\cos \theta } \\ \ & = \frac{1}{\cos \theta }+ \frac{\sin \theta }{ \cos \theta } \\ \ & = \sec \theta + \tan \theta \\ \ & = \text{RHS}\\ \end{align}
6. (j) Prove that: \( \frac{1}{\sec A - \tan A} - \frac{1}{\cos A } = \frac{1}{\cos A} - \frac{1} {\sec A + \tan A } \)
Solution: By transposition, \( \frac{1}{\sec A - \tan A} + \frac{1} {\sec A + \tan A } = \frac{1}{\cos A} + \frac{1}{\cos A} \) \begin{align} \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^2 A - \tan^2 A}\\ \ & = \frac{2\sec A }{\sec^2 A - \tan^2 A}\\ \ & = \frac{2\sec A }{1} \\ \ & = 2\sec A \\ \ & = \sec A + \sec A \\ \ & = \frac{1}{\cos A} + \frac{1}{\cos A} \\ \ & = \text{New RHS}\\ \end{align}
6. (k) Prove that: \( \frac{1-\sec A + \tan A}{1+\sec A - \tan A} = \frac{ \sec A + \tan A - 1}{ \sec A + \tan A + 1} \)
Solution: \begin{align} \text{LHS} & = \frac{1-\sec A + \tan A}{1+\sec A - \tan A}\\ \ & = \frac{(\sec^2 A - \tan^2 A) - (\sec A - \tan A) }{(\sec^2 A - \tan^2 A )+(\sec A - \tan A)}\\ \ & = \frac{(\sec A - \tan A ) ( \sec A + \tan A ) -(\sec A - \tan A) }{(\sec A - \tan A )(\sec A + \tan A ) + (\sec A - \tan A) }\\ & = \frac{(\sec A - \tan A) ( \sec A + \tan A -1) }{(\sec A - \tan A ) ( \sec A + \tan A + 1)}\\ \ & = \frac{ \sec A + \tan A - 1}{ \sec A + \tan A + 1} \\ \ & = \text{RHS}\\ \end{align}
6. (l) Prove that: \( \frac{\cot A + \text{cosec} A - 1}{\cot A + \text{cosec} A + 1} = \frac{ 1-\text{cosec} A - \cot A } {1+\text{cosec} A - \cot A } \)
Solution: \begin{align} \text{LHS} & = \frac{\cot A + \text{cosec} A - 1}{\cot A + \text{cosec} A + 1}\\ \ & = \frac{ ( \text{cosec} A + \cot A) - ( \text{cosec}^2 A - \cot^2 A) }{ ( \text{cosec} A + \cot A) + ( \text{cosec}^2 A - \cot^2 A) }\\ \ & = \frac{ ( \text{cosec} A + \cot A) - ( \text{cosec} A - \cot A)( \text{cosec} A + \cot A) }{ ( \text{cosec} A + \cot A) + ( \text{cosec} A - \cot A)( \text{cosec} A - \cot A) }\\ \ & = \frac{ (\text{cosec A} + \cot A) [1-(\text{cosec} A + \cot A)] } {(\text{cosec A} + \cot A) [1-(\text{cosec} A - \cot A)]}\\ \ & = \frac{ 1-\text{cosec} A - \cot A } {1+\text{cosec} A + \cot A }\\ \ & = \text{RHS}\\ \end{align}
6. (m) Prove that: \( ( 1+ \sin A + \cos A)^2 = 2(1+\sin A)(1+\cos A) \)
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}
6. (n ) Prove that: \( ( 1-\sin \alpha - \cos \alpha)^2 = 2(1-\sin \alpha) ( 1-\cos \alpha )\)
Solution: \begin{align} \text{LHS} & = ( 1-\sin \alpha - \cos \alpha)^2 \\ \ & = [( 1-\sin \alpha) - \cos \alpha ]^2 \\ \ & = (1-\sin \alpha)^2 -2\times (1-\sin \alpha ) \times \cos \alpha + \cos^2 \alpha \\ \ & = (1-\sin \alpha)^2 -2\times (1-\sin \alpha ) \times \cos \alpha + 1-\sin^2 \alpha \\ \ & = (1-\sin \alpha)^2 -2\times (1-\sin \alpha ) \times \cos \alpha + (1-\sin \alpha)(1+\sin \alpha ) \\ \ & = (1-\sin \alpha) (1-\sin \alpha -2\cos \alpha + 1+\sin \alpha )\\ \ & = (1-\sin \alpha) ( 2-2\cos \alpha )\\ \ & = (1-\sin \alpha)\times 2 ( 1-\cos \alpha )\\ \ & = 2(1-\sin \alpha) ( 1-\cos \alpha )\\ \ & = \text{RHS}\\ \end{align}
6. (0) Prove that: \( \sin^2 x \times \sec^2 x + \tan^2 x \times \cos^2 x = \sin^2 x + \tan^2 x \)
Solution: \begin{align} \text{LHS} & = \sin^2 x \times \sec^2 x + \tan^2 x \times \cos^2 x\\ \ & = \sin^2 x \times (1+\tan^2 x) + \tan^2 x \times (1-\sin^2 x)\\ \ & = \sin^2 x + \sin^2 x \tan^2 x + \tan^2 x - \sin^2 x \tan^2 x \\ \ & = \sin^2 x + \tan^2 x \\ \ & = \text{RHS}\\ \end{align}
6. (p) Prove that: \( \frac{\tan x }{\sec x - 1} - \frac{ \sin x }{1+ \cos x} = 2\cot x \)
Solution: \begin{align} \text{LHS} & = \frac{\tan x }{\sec x - 1} - \frac{ \sin x }{1+ \cos x}\\ \ & = \frac{\dfrac{\sin x}{\cos x} }{\dfrac{1}{\cos x} - 1} - \frac{ \sin x }{1+ \cos x}\\ \ & = \frac{\dfrac{\sin x}{\cos x} }{\dfrac{1-\cos x}{\cos x} } - \frac{ \sin x }{1+ \cos x}\\ \ & = \frac{\sin x } {1-\cos x} - \frac{\sin x}{1+\cos x} \\ \ & = \frac{\sin x ( 1+ \cos x) -\sin x ( 1- \cos x)}{(1-\cos x)( 1+\cos x )}\\ \ & = \frac{\sin x + \sin x \cos x - sin x + \sin x \cos x }{ 1^2 - \cos^2 x }\\ \ & = \frac{2\sin x \cos x }{\sin^2 x }\\ \ & = \frac{2\cos x }{\sin x}\\ \ & = 2\cot x \\ \ & = \text{RHS}\\ \end{align}
6. (q) Prove that: \( \frac{\cos A - \sin A + 1}{\cos A + \sin A -1} = \frac{1+\cos A }{ \sin A } \)
Solution: \begin{align} \text{LHS} & = \frac{\cos A - \sin A + 1}{\cos A + \sin A -1} \\ \ & = \frac{\dfrac{\cos A - \sin A + 1}{\sin A}}{\dfrac{\cos A + \sin A -1}{\sin A} } \\ \ & = \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{\text{cosec} A + \cot A - 1}{1-\text{cosec}A + \cot A} \\ \ & = \frac{(\text{cosec} A + \cot A) -(\text{cosec}^2 A + \cot^2 A) }{1-\text{cosec}A + \cot A} \\ \ & = \frac{(\text{cosec} A + \cot A) -(\text{cosec} A - \cot A)(\text{cosec} A + \cot A) }{1-\text{cosec}A + \cot A} \\ \ & = \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}
6. (r) Prove that: \( \frac{1}{\cot A ( 1- \cot A )} + \frac{1}{\tan A (1-\tan A )} = 1+ \sec A \text{cosec} A \)
Solution: \begin{align} \text{LHS} & = \frac{1}{\cot A ( 1- \cot A )} + \frac{1}{\tan A (1-\tan A )} \\ \ & = \frac{1}{\dfrac{\cos A}{\sin A} \left( 1- \dfrac{\cos A}{\sin A} \right)} + \frac{1}{\dfrac{\sin A}{\cos A} \left(1- \dfrac{\sin A}{\cos A} \right)} \\ \ & = \frac{ \sin A}{\cos A \times \dfrac{\sin A - \cos A}{\sin A} }+ \frac{ \cos A}{\sin A \times \dfrac{\cos A - \sin A }{\cos A}} \\ \ & = \frac{ \sin^2 A}{\cos A ( \sin A - \cos A)} + \frac{\cos^2 A}{\sin A( \cos A - \sin A )} \\ \ & = \frac{ \sin^2 A}{\cos A ( \sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A )} \\ \ & = \frac{\sin^3 A - \cos^3 A }{\sin A \cos 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 A - \cos A ) }\\ \ & = \frac{ \sin^2 A + \cos^2 A+ \sin A \cos A }{\sin A \cos A }\\ \ & = \frac{ 1+ \sin A \cos A }{\sin A \cos A }\\ \ & = \frac{ 1 }{\sin A \cos A }+ \frac{\sin A \cos A}{\sin A \cos A}\\ \ & = \sec A \text{ cosec} A + 1\\ \ & = 1+ \sec A \text{ cosec} A \\ \ & = \text{RHS} \end{align} Alternative, \begin{align} \text{LHS} & = \frac{1}{\cot A ( 1- \cot A )} + \frac{1}{\tan A (1-\tan A )} \\ \ & = \frac{1}{\dfrac{1}{\tan A} \left( 1- \dfrac{1}{\tan A} \right)} + \frac{1}{\tan A (1-\tan A )} \\ \ & = \frac{\tan A}{\dfrac{\tan A - 1}{\tan A}} + \frac{1}{\tan A (1-\tan A )} \\ \ & = \frac{\tan^2 A}{\tan A -1 } - \frac{1}{\tan A (\tan A - 1)} \\ \ & = \frac{\tan^3 A - 1}{\tan A (\tan A - 1 ) } \\ \ & = \frac{\tan^3 A - 1^3}{\tan A (\tan A - 1 ) } \\ \ & = \frac{(\tan A - 1)(\tan^2 A + \tan A \times 1 + 1^2 )}{\tan A (\tan A - 1 ) } \\ \ & = \frac{\tan^2 A + \tan A + 1}{\tan A }\\ \ & = \frac{\tan^2 A }{\tan A} + \frac{\tan A}{\tan A} + \frac{1}{\tan A} \\ \ & = \tan A + 1 + \cot A \\ \ & =1+ \tan A + \cot A \\ \ & = 1+ \frac{\sin A}{\cos A} + \frac{\cos A }{\sin A} \\ \ & = 1+ \frac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\ \ & = 1+ \frac{1}{\cos A \sin A} \\ \ & = 1+ \sec A \text{cosec} A \\ \ & = \text{RHS}\\ \end{align}
6. (s) Prove that: \( \frac{\sec A - \tan A }{ \sec A + \tan A } = 1 -2\sec A \tan A + 2\tan^2 A \)
Solution: \begin{align} \text{LHS} & = \frac{\sec A - \tan A }{ \sec A + \tan A } \\ \ & = \frac{\sec A - \tan A }{ \sec A + \tan A } \times \frac{\sec A - \tan A }{ \sec A - \tan A } \\ \ & = \frac{ ( \sec A - \tan A)^2 }{\sec^2 A - \tan^2 A } \\ \ & = \frac{ \sec^2 - 2\sec A \tan A + \tan^2 A }{1} \\ \ & = 1+\tan^2 A - 2\sec A \tan A + \tan^2 A \\ \ & = 1 -2\sec A \tan A + 2\tan^2 A\\ \ & = \text{RHS}\\ \end{align}
6. (t) Prove that: \( \frac{\cos A - \sin A + 1}{ \cos A + \sin A + 1 } = \frac{ 1 -\sin A}{\cos A} \)
Solution: \begin{align} \text{LHS} & =\frac{\cos A - \sin A + 1}{ \cos A + \sin A + 1} \\ \ & =\frac{\dfrac{\cos A - \sin A + 1}{\cos A}}{ \dfrac{\cos A + \sin A + 1}{\cos A}} \\ \ & = \frac{\dfrac{\cos A}{\cos A} - \dfrac{\sin A}{\cos A} + \dfrac{1}{\cos A}} {\dfrac{\cos A}{\cos A}+\dfrac{\sin A}{\cos A} + \dfrac{1}{\cos A}} \\ \ & = \frac{1-\tan A + \sec A}{1+\tan A + \sec A } \\ \ & = \frac{\sec^2 A -\tan^2 A + (\sec A - \tan 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)(\sec A + \tan A +1) }{1+\tan A + \sec A } \\ \ & = \frac{(\sec A -\tan A ) ( \sec A + \tan A + 1) } { 1 + \sec A + \tan A } \\ \ & = \sec A -\tan A \\ \ & = \frac{1}{\cos A} - \frac{\sin A}{\cos A} \\ \ & = \frac{ 1 -\sin A}{\cos A} \\ \ & = \text{RHS}\\ \end{align}
6. (u ) Prove that: \( \frac{\sin \theta + \cos \theta + 1 }{ \sin \theta + \cos \theta - 1 } - \frac{ 1+ \sin \theta - \cos \theta }{ 1-\sin \theta + \cos \theta } = 2(1+\text{cosec} \theta ) \)
Solution: \begin{align} \text{LHS} \ & = \frac{\sin \theta + \cos \theta +1 }{\sin \theta + \cos \theta - 1} - \frac{1+\sin \theta -\cos \theta }{1-\sin \theta + \cos \theta } \\ \ & =\frac{(\sin \theta +\cos \theta +1 )(1-\sin \theta +\cos \theta ) - (1+\sin \theta - \cos \theta )(\sin \theta + \cos \theta -1)}{(\sin \theta + \cos \theta -1)(1- \sin \theta + \cos \theta ) } \\ \ & =\frac{\{(\cos \theta + 1)+\sin \theta \}\{(\cos \theta + 1)-\sin \theta \} -\{\sin \theta +(1 - \cos \theta )\}\{\sin \theta -(1- \cos \theta )\}}{\{\cos \theta+ (\sin \theta -1) \}\{\cos \theta - (\sin \theta -1) \}} \\ \ & = \frac{(\cos \theta +1)^2 - \sin^2 \theta - [ \sin^2 \theta - (1-\cos \theta)^2 ]}{\cos^2 \theta - (\sin \theta -1)^2 }\\ \ & = \frac{(\cos \theta +1)^2 - \sin^2 \theta - \sin^2 \theta + (1-\cos \theta)^2 }{\cos^2 \theta - (\sin \theta -1)^2 }\\ \ & = \frac{\cos^2 \theta + 2\cos \theta +1 -2\sin^2 \theta + 1 - 2\cos \theta + \cos^2 \theta }{\cos^2 \theta - ( \sin^2 \theta - 2\sin \theta +1 )} \\ \ & = \frac{2+ 2\cos^2 \theta-2\sin^2 \theta }{\cos^2 \theta -\sin^2 \theta + 2\sin \theta -1}\\ \ & = \frac{ 2+2(1-\sin^2 \theta ) -2\sin^2 \theta }{1-\sin^2 \theta -\sin^2 \theta + 2\sin \theta -1 }\\ \ & = \frac{2+2-2\sin^2 \theta -2\sin^2 \theta }{2\sin \theta - 2\sin^2 \theta }\\ \ & = \frac{ 4 - 4\sin^2 \theta }{2\sin \theta ( 1-\sin \theta ) }\\ \ & = \frac{4(1-\sin^2 \theta )}{2\sin \theta ( 1-\sin \theta )}\\ \ & = \frac{4(1-\sin^2 \theta )}{2\sin \theta ( 1-\sin \theta ) } \\ \ & = \frac{2(1^2-\sin^2 \theta )}{\sin \theta ( 1-\sin \theta ) } \\ \ & = \frac{2(1-\sin \theta )(1+\sin \theta )}{\sin \theta ( 1-\sin \theta ) } \\ \ & = \frac{2 (1+\sin \theta )}{\sin \theta } \\ \ & = 2\left(\frac{1}{\sin \theta } + \frac{\sin \theta }{\sin \theta } \right) \\ \ & = 2 ( \text{cosec } \theta + 1 )\\ \ & = 2 (1 + \text{cosec } \theta )\\ \ & = \text {RHS} \\ \end{align}
6. (v) Prove that: \( \text{cosec}^4 A (1-\cos^4 A ) = 1 + 2\cot^2 A \)
Solution: \begin{align} \text{LHS} & = \text{cosec}^4 A (1-\cos^4 A )\\ \ & = \frac{1}{\sin^4 A} \times [1^2 - (\cos^2 A)^2 ] \\ \ & = \frac{1}{\sin^4 A} \times (1-\cos^2 A)(1+\cos^2 A) \\ \ & = \frac{1}{\sin^4 A} \times (\sin^2 A) ( 1+\cos^2 A ) \\ \ & = \frac{1+\cos^2 A}{\sin^2 A }\\ \ & = \frac{1}{\sin^2 A} + \frac{\cos^2 A}{\sin^2 A}\\ \ & = \text{cosec}^2 A + \cot^2 A \\ \ & = 1+\cot^2 A + \cot^2 A \\ \ & = 1 + 2\cot^2 A\\ \ & = \text{RHS} \end{align}
6. (w) Prove that: \( (3-4\sin^2 x)( \sec^2 x - 4\tan^2 x) = (3-\tan^2 x)(1-4\sin^2 x) \)
Solution: \begin{align} \text{LHS} & = (3-4\sin^2 x)( \sec^2 x - 4\tan^2 x)\\ \ & = (3-4\sin^2 x) \left( \frac{1}{\cos^2 x} - 4 \times \frac{\sin^2 x}{\cos^2 x} \right)\\ \ & = (3-4\sin^2 x) \left( \frac{1-4\sin^2 x}{\cos^2 x}\right )\\ \ & = \left(\frac{3-4\sin^2 x}{\cos^2 x } \right) (1-4\sin^2 x )\\ \ & = \left( \frac{3}{\cos^2 x} - \frac{4\sin^2 x}{\cos^2 x} \right) (1-4\sin^2 x ) \\ \ & = ( 3\sec^2 x -4\tan^2 x) (1-4\sin^2 x ) \\ \ & = [ 3(1+\tan^2 x) -4\tan^2 x] (1-4\sin^2 x ) \\ \ & = (3+3\tan^2 x -4\tan^2 x) (1-4\sin^2 x ) \\ \ & = (3- \tan^2 x)(1-4\sin^2 x) \\ \ & = \text{RHS} \end{align}
6. (x) Prove that: \( (sec A + \text{cosec} A)^2 = (1+\cot A)^2 + (1+\tan A)^2 \)
Solution: \begin{align} \text{LHS} & = (\sec A + \text{cosec} A)^2\\ \ & = \sec^2 A + 2\sec A \text{cosec} A +\text{cosec}^2 A\\ \ & = \sec^2 A+ 2\times \frac{1}{\cos A} \times \frac{1}{\sin A} +\text{cosec}^2 A\\ \ & = \sec^2 A+ 2\times \frac{\sin^2 A + \cos^2 A}{\cos A \sin A} +\text{cosec}^2 A\\ \ & = \sec^2 A+ 2\left( \frac{\sin^2 A}{\cos A \sin A} + \frac{\cos^2 A}{\cos A \sin A}\right) +\text{cosec}^2 A\\ \ & = \sec^2 A+ 2\left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \right) +\text{cosec}^2 A\\ \ & = 1+ \tan^2 A + 2(\tan A +\cot A) + 1+\cot^2 A \\ \ & = 1 + \tan^2 A + 2\tan A + 2\cot A + 1 +\cot^2 A \\ \ & = 1+2\tan A + \tan^2 A + 1+ 2\cot A + \cot^2 A \\ \ & = 1^2 + 2\times 1 \times \tan A + \tan^2 A + 1^2 + 2\times 1 \times \cot A + \cot^2 A \\ \ & = (1+\cot A)^2 + (1+\tan A)^2 \\ \ & = \text{RHS} \end{align}
7. (a) Prove that: \( \sqrt{\frac{\sec \theta +1}{\sec \theta - 1}} = \frac{1+\cos \theta }{\sin \theta } \)
Solution: \begin{align} \text{LHS} & = \sqrt{\frac{\sec \theta +1}{\sec \theta - 1}} \\ \ & = \sqrt{\frac{\sec \theta +1}{\sec \theta - 1}\times \frac{\sec \theta + 1}{\sec \theta +1}} \\ \ & = \sqrt{\frac{( \sec \theta +1)^2 }{\sec^2 \theta - 1^2}} \\ \ & = \frac{\sec \theta +1}{\sqrt{\sec^2 \theta - 1}} \\ \ & = \frac{\sec \theta +1}{\tan \theta } \\ \ & = \dfrac{\dfrac{1}{\cos \theta} +1}{\dfrac{\sin \theta}{\cos \theta }}\\ \ & = \dfrac{\dfrac{1+\cos \theta }{\cos \theta} }{\dfrac{\sin \theta }{\cos \theta }}\\ \ & = \frac{1+\cos \theta }{\sin \theta } \\ \ & = \text{RHS} \end{align}
7. (b) Prove that: \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta} } = \sec \theta + \tan \theta \)
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}{1^2-\sin^2 \theta} } \\ \ & = \frac{1+\sin \theta }{\sqrt{1-\sin^2 \theta }} \\ \ & = \frac{1+\sin \theta }{\sqrt{\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}
7. (c) Prove that: \( \sqrt{\frac{1-\cos A}{1+\cos A }} = \text{cosec} A - \cot A \)
Solution: \begin{align} \text{LHS} & = \sqrt{\frac{1-\cos A}{1+\cos A }}\\ \ & = \sqrt{\frac{1-\cos A}{1+\cos A }\times \frac{1-\cos A}{1-\cos A}}\\ \ & = \sqrt{\frac{(1-\cos A)^2}{1^2 - \cos^2 A }}\\ \ & = \frac{1- \cos A }{\sqrt{1-\cos^2 A }} \\ \ & = \frac{1-\cos A}{\sin A} \\ \ & = \frac{1}{\sin A} - \frac{\cos A}{\sin A} \\ \ & = \text{cosec} A - \cot A \\ \ & = \text{RHS} \end{align}
7. (d) Prove that: \( \sqrt{\frac{1+\sin A}{1-\sin A}} - \sqrt{\frac{1-\sin A }{1+\sin A}} = 2\tan A\)
Solution: \begin{align} \text{LHS} & = \sqrt{\frac{1+\sin A}{1-\sin A}} - \sqrt{\frac{1-\sin A }{1+\sin A}} \\ \ & = \frac{\sqrt{1+\sin A}}{\sqrt{1-\sin A}} - \frac{\sqrt{1-\sin A}}{\sqrt{1+\sin A}} \\ \ & = \frac{(\sqrt{1+\sin A})^2-(\sqrt{1-\sin A} )^2}{\sqrt{1-\sin A} \sqrt{1+\sin A}} \\ \ & = \frac{1+\sin A - (1-\sin A) }{\sqrt{1^2 -\sin^2 A} }\\ \ & = \frac{ 1+\sin A - 1 + \sin A } {\sqrt{1-\sin^2 A }} \\ \ & = \frac{2\sin A} {\cos A} \\ \ & = 2\tan A \\ \ & = \text{RHS} \end{align}
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