\displaystyle \textbf{Question 1: }\text{Prove the following trigonometric identity:}
\displaystyle (1-\cos^2 A)\mathrm{cosec}^2 A=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1-\cos^2 A)\mathrm{cosec}^2 A
\displaystyle =\sin^2 A\,\mathrm{cosec}^2 A\qquad[\because 1-\cos^2 A=\sin^2 A]
\displaystyle =\sin^2 A\times\frac{1}{\sin^2 A}
\displaystyle =1=\text{RHS}
\displaystyle \therefore (1-\cos^2 A)\mathrm{cosec}^2 A=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove the following trigonometric identity:}
\displaystyle (1+\cot^2 A)\sin^2 A=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\cot^2 A)\sin^2 A
\displaystyle =\mathrm{cosec}^2 A\sin^2 A\qquad[\because 1+\cot^2 A=\mathrm{cosec}^2 A]
\displaystyle =\frac{1}{\sin^2 A}\times\sin^2 A
\displaystyle =1=\text{RHS}
\displaystyle \therefore (1+\cot^2 A)\sin^2 A=1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove the following trigonometric identity:}
\displaystyle (\sec^2\theta-1)(\mathrm{cosec}^2\theta-1)=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sec^2\theta-1)(\mathrm{cosec}^2\theta-1)
\displaystyle =\tan^2\theta\cot^2\theta
\displaystyle \qquad[\because \sec^2\theta-1=\tan^2\theta\text{ and }\mathrm{cosec}^2\theta-1=\cot^2\theta]
\displaystyle =(\tan\theta\cot\theta)^2
\displaystyle =1=\text{RHS}
\displaystyle \therefore (\sec^2\theta-1)(\mathrm{cosec}^2\theta-1)=1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove the following trigonometric identity:}
\displaystyle \tan\theta+\frac{1}{\tan\theta}=\sec\theta\,\mathrm{cosec}\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan\theta+\frac{1}{\tan\theta}
\displaystyle =\tan\theta+\cot\theta
\displaystyle =\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}
\displaystyle =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{1}{\sin\theta\cos\theta}\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle =\sec\theta\,\mathrm{cosec}\theta=\text{RHS}
\displaystyle \therefore \tan\theta+\frac{1}{\tan\theta}=\sec\theta\,\mathrm{cosec}\theta.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos\theta}{1+\sin\theta}
\displaystyle =\frac{\cos\theta}{1+\sin\theta}\times\frac{1-\sin\theta}{1-\sin\theta}
\displaystyle =\frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta}
\displaystyle =\frac{\cos\theta(1-\sin\theta)}{\cos^2\theta}\qquad[\because 1-\sin^2\theta=\cos^2\theta]
\displaystyle =\frac{1-\sin\theta}{\cos\theta}=\text{RHS}
\displaystyle \therefore \frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove the following trigonometric identity:}
\displaystyle \cos^2 A+\frac{1}{1+\cot^2 A}=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^2 A+\frac{1}{1+\cot^2 A}
\displaystyle =\cos^2 A+\frac{1}{\mathrm{cosec}^2 A}\qquad[\because 1+\cot^2 A=\mathrm{cosec}^2 A]
\displaystyle =\cos^2 A+\sin^2 A
\displaystyle =1=\text{RHS}\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle \therefore \cos^2 A+\frac{1}{1+\cot^2 A}=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1-\cos\theta}{\sin\theta}
\displaystyle =\frac{1-\cos\theta}{\sin\theta}\times\frac{1+\cos\theta}{1+\cos\theta}
\displaystyle =\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)}
\displaystyle =\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}\qquad[\because 1-\cos^2\theta=\sin^2\theta]
\displaystyle =\frac{\sin\theta}{1+\cos\theta}=\text{RHS}
\displaystyle \therefore \frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\sin\theta}{1-\cos\theta}=\mathrm{cosec}\theta+\cot\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin\theta}{1-\cos\theta}
\displaystyle =\frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}
\displaystyle =\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}
\displaystyle =\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}\qquad[\because 1-\cos^2\theta=\sin^2\theta]
\displaystyle =\frac{1+\cos\theta}{\sin\theta}
\displaystyle =\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}
\displaystyle =\mathrm{cosec}\theta+\cot\theta=\text{RHS}
\displaystyle \therefore \frac{\sin\theta}{1-\cos\theta}=\mathrm{cosec}\theta+\cot\theta.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1-\sin\theta}{1+\sin\theta}=(\sec\theta-\tan\theta)^2
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=(\sec\theta-\tan\theta)^2
\displaystyle =\left(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}\right)^2
\displaystyle =\left(\frac{1-\sin\theta}{\cos\theta}\right)^2
\displaystyle =\frac{(1-\sin\theta)^2}{\cos^2\theta}
\displaystyle =\frac{(1-\sin\theta)^2}{1-\sin^2\theta}
\displaystyle =\frac{(1-\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}
\displaystyle =\frac{1-\sin\theta}{1+\sin\theta}=\text{LHS}
\displaystyle \therefore \frac{1-\sin\theta}{1+\sin\theta}=(\sec\theta-\tan\theta)^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}
\displaystyle =\frac{\mathrm{cosec}^2\theta\,\tan\theta}{\sec^2\theta}
\displaystyle =\frac{\frac{1}{\sin^2\theta}\times\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}
\displaystyle =\frac{1}{\sin\theta\cos\theta}\times\cos^2\theta
\displaystyle =\frac{\cos\theta}{\sin\theta}
\displaystyle =\cot\theta=\text{RHS}
\displaystyle \therefore \frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove the following trigonometric identity:}
\displaystyle \tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan^2\theta-\sin^2\theta
\displaystyle =\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta
\displaystyle =\sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right)
\displaystyle =\sin^2\theta(\sec^2\theta-1)
\displaystyle =\sin^2\theta\tan^2\theta\qquad[\because \sec^2\theta-1=\tan^2\theta]
\displaystyle =\tan^2\theta\sin^2\theta=\text{RHS}
\displaystyle \therefore \tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Prove the following trigonometric identity:}
\displaystyle (1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)
\displaystyle =\sec^2\theta(1-\sin^2\theta)
\displaystyle =\sec^2\theta\cos^2\theta\qquad[\because 1-\sin^2\theta=\cos^2\theta]
\displaystyle =\frac{1}{\cos^2\theta}\times\cos^2\theta
\displaystyle =1=\text{RHS}
\displaystyle \therefore (1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)=1.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove the following trigonometric identity:}
\displaystyle \sin^2 A\cot^2 A+\cos^2 A\tan^2 A=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^2 A\cot^2 A+\cos^2 A\tan^2 A
\displaystyle =\sin^2 A\left(\frac{\cos^2 A}{\sin^2 A}\right)+\cos^2 A\left(\frac{\sin^2 A}{\cos^2 A}\right)
\displaystyle =\cos^2 A+\sin^2 A
\displaystyle =1=\text{RHS}\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle \therefore \sin^2 A\cot^2 A+\cos^2 A\tan^2 A=1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}
\displaystyle =\frac{(1+\sin\theta)^2+\cos^2\theta}{\cos\theta(1+\sin\theta)}
\displaystyle =\frac{1+2\sin\theta+\sin^2\theta+\cos^2\theta}{\cos\theta(1+\sin\theta)}
\displaystyle =\frac{2+2\sin\theta}{\cos\theta(1+\sin\theta)}\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle =\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}
\displaystyle =\frac{2}{\cos\theta}
\displaystyle =2\sec\theta=\text{RHS}
\displaystyle \therefore \frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{(1+\sin\theta)^2+(1-\sin\theta)^2}{2\cos^2\theta}=\frac{1+\sin^2\theta}{1-\sin^2\theta}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(1+\sin\theta)^2+(1-\sin\theta)^2}{2\cos^2\theta}
\displaystyle =\frac{1+2\sin\theta+\sin^2\theta+1-2\sin\theta+\sin^2\theta}{2\cos^2\theta}
\displaystyle =\frac{2+2\sin^2\theta}{2\cos^2\theta}
\displaystyle =\frac{1+\sin^2\theta}{\cos^2\theta}
\displaystyle =\frac{1+\sin^2\theta}{1-\sin^2\theta}\qquad[\because \cos^2\theta=1-\sin^2\theta]
\displaystyle =\text{RHS}
\displaystyle \therefore \frac{(1+\sin\theta)^2+(1-\sin\theta)^2}{2\cos^2\theta}=\frac{1+\sin^2\theta}{1-\sin^2\theta}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1+\tan^2\theta}{1+\cot^2\theta}=\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2=\tan^2\theta
\displaystyle \text{Answer:}
\displaystyle \frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\mathrm{cosec}^2\theta}
\displaystyle =\frac{\frac{1}{\cos^2\theta}}{\frac{1}{\sin^2\theta}}
\displaystyle =\frac{\sin^2\theta}{\cos^2\theta}
\displaystyle =\tan^2\theta
\displaystyle \text{Also, }\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2
\displaystyle =\left(\frac{1-\tan\theta}{1-\frac{1}{\tan\theta}}\right)^2
\displaystyle =\left(\frac{1-\tan\theta}{\frac{\tan\theta-1}{\tan\theta}}\right)^2
\displaystyle =\left(\frac{\tan\theta(1-\tan\theta)}{-(1-\tan\theta)}\right)^2
\displaystyle =(-\tan\theta)^2
\displaystyle =\tan^2\theta
\displaystyle \therefore \frac{1+\tan^2\theta}{1+\cot^2\theta}=\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2=\tan^2\theta.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }\frac{1+\sec\theta}{\sec\theta}=\frac{\sin^2\theta}{1-\cos\theta}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\sec\theta}{\sec\theta}
\displaystyle =\frac{1}{\sec\theta}+1
\displaystyle =\cos\theta+1
\displaystyle =1+\cos\theta
\displaystyle \text{RHS}=\frac{\sin^2\theta}{1-\cos\theta}
\displaystyle =\frac{1-\cos^2\theta}{1-\cos\theta}\qquad[\because \sin^2\theta=1-\cos^2\theta]
\displaystyle =\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}
\displaystyle =1+\cos\theta
\displaystyle \therefore \text{LHS}=\text{RHS}.

\displaystyle \text{(ii) }\frac{1+\mathrm{cosec}\theta}{\mathrm{cosec}\theta}=\frac{\cos^2\theta}{1-\sin\theta}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\mathrm{cosec}\theta}{\mathrm{cosec}\theta}
\displaystyle =\frac{1}{\mathrm{cosec}\theta}+1
\displaystyle =\sin\theta+1
\displaystyle =1+\sin\theta
\displaystyle \text{RHS}=\frac{\cos^2\theta}{1-\sin\theta}
\displaystyle =\frac{1-\sin^2\theta}{1-\sin\theta}\qquad[\because \cos^2\theta=1-\sin^2\theta]
\displaystyle =\frac{(1-\sin\theta)(1+\sin\theta)}{1-\sin\theta}
\displaystyle =1+\sin\theta
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove the following trigonometric identity:}
\displaystyle \sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\tan^6\theta+3\tan^2\theta\sec^2\theta+1
\displaystyle =\tan^6\theta+3\tan^2\theta(1+\tan^2\theta)+1
\displaystyle \qquad[\because \sec^2\theta=1+\tan^2\theta]
\displaystyle =\tan^6\theta+3\tan^4\theta+3\tan^2\theta+1
\displaystyle =(1+\tan^2\theta)^3
\displaystyle =(\sec^2\theta)^3
\displaystyle =\sec^6\theta=\text{LHS}
\displaystyle \therefore \sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\cos^2 A}{(1+\sin A)^2}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sec A-\tan A}{\sec A+\tan A}
\displaystyle =\frac{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}
\displaystyle =\frac{1-\sin A}{1+\sin A}
\displaystyle =\frac{(1-\sin A)(1+\sin A)}{(1+\sin A)^2}
\displaystyle =\frac{1-\sin^2 A}{(1+\sin A)^2}
\displaystyle =\frac{\cos^2 A}{(1+\sin A)^2}\qquad[\because 1-\sin^2 A=\cos^2 A]
\displaystyle =\text{RHS}
\displaystyle \therefore \frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\cos^2 A}{(1+\sin A)^2}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Prove the following trigonometric identity:}
\displaystyle (\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\tan^2\theta+\sin^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)
\displaystyle =\sec^2\theta-\cos^2\theta
\displaystyle =(1+\tan^2\theta)-\cos^2\theta
\displaystyle =\tan^2\theta+(1-\cos^2\theta)
\displaystyle =\tan^2\theta+\sin^2\theta\qquad[\because 1-\cos^2\theta=\sin^2\theta]
\displaystyle =\text{RHS}
\displaystyle \therefore (\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\tan^2\theta+\sin^2\theta.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1+\sin A}{1-\sin A}}
\displaystyle =\sqrt{\frac{1+\sin A}{1-\sin A}\times\frac{1+\sin A}{1+\sin A}}
\displaystyle =\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}}
\displaystyle =\sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}\qquad[\because 1-\sin^2 A=\cos^2 A]
\displaystyle =\frac{1+\sin A}{\cos A}
\displaystyle =\frac{1}{\cos A}+\frac{\sin A}{\cos A}
\displaystyle =\sec A+\tan A=\text{RHS}
\displaystyle \therefore \sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A.

\displaystyle \text{(ii) }\sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2\mathrm{cosec} A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}
\displaystyle =\sqrt{\frac{(1-\cos A)^2}{1-\cos^2 A}}+\sqrt{\frac{(1+\cos A)^2}{1-\cos^2 A}}
\displaystyle =\frac{1-\cos A}{\sin A}+\frac{1+\cos A}{\sin A}\qquad[\because 1-\cos^2 A=\sin^2 A]
\displaystyle =\frac{2}{\sin A}
\displaystyle =2\mathrm{cosec} A=\text{RHS}
\displaystyle \therefore \sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2\mathrm{cosec} A.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }\sqrt{\frac{\sec\theta-1}{\sec\theta+1}}+\sqrt{\frac{\sec\theta+1}{\sec\theta-1}}=2\mathrm{cosec}\theta
\displaystyle \hfill\text{[CBSE 2001, 2006C, 2023, 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{\sec\theta-1}{\sec\theta+1}}+\sqrt{\frac{\sec\theta+1}{\sec\theta-1}}
\displaystyle =\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}+\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}
\displaystyle =\sqrt{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}}+\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}
\displaystyle =\frac{1-\cos\theta}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}
\displaystyle \qquad[\because 1-\cos^2\theta=\sin^2\theta]
\displaystyle =\frac{2}{\sin\theta}
\displaystyle =2\mathrm{cosec}\theta=\text{RHS}
\displaystyle \therefore \sqrt{\frac{\sec\theta-1}{\sec\theta+1}}+\sqrt{\frac{\sec\theta+1}{\sec\theta-1}}=2\mathrm{cosec}\theta.

\displaystyle \text{(ii) }\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=2\sec\theta
\displaystyle \hfill\text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}
\displaystyle =\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}+\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}
\displaystyle =\frac{1+\sin\theta}{\cos\theta}+\frac{1-\sin\theta}{\cos\theta}
\displaystyle \qquad[\because 1-\sin^2\theta=\cos^2\theta]
\displaystyle =\frac{2}{\cos\theta}
\displaystyle =2\sec\theta=\text{RHS}
\displaystyle \therefore \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=2\sec\theta.

\displaystyle \text{(iii) }\sqrt{\frac{\mathrm{cosec}\theta-1}{\mathrm{cosec}\theta+1}}+\sqrt{\frac{\mathrm{cosec}\theta+1}{\mathrm{cosec}\theta-1}}=2\sec\theta
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{\mathrm{cosec}\theta-1}{\mathrm{cosec}\theta+1}}+\sqrt{\frac{\mathrm{cosec}\theta+1}{\mathrm{cosec}\theta-1}}
\displaystyle =\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}+\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}
\displaystyle =\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}+\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}
\displaystyle =\frac{1-\sin\theta}{\cos\theta}+\frac{1+\sin\theta}{\cos\theta}
\displaystyle \qquad[\because 1-\sin^2\theta=\cos^2\theta]
\displaystyle =\frac{2}{\cos\theta}
\displaystyle =2\sec\theta=\text{RHS}
\displaystyle \therefore \sqrt{\frac{\mathrm{cosec}\theta-1}{\mathrm{cosec}\theta+1}}+\sqrt{\frac{\mathrm{cosec}\theta+1}{\mathrm{cosec}\theta-1}}=2\sec\theta.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\tan^2 A}{1+\tan^2 A}+\frac{\cot^2 A}{1+\cot^2 A}=1\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\tan^2 A}{1+\tan^2 A}+\frac{\cot^2 A}{1+\cot^2 A}
\displaystyle =\frac{\tan^2 A}{\sec^2 A}+\frac{\cot^2 A}{\mathrm{cosec}^2 A}
\displaystyle \qquad[\because 1+\tan^2 A=\sec^2 A\text{ and }1+\cot^2 A=\mathrm{cosec}^2 A]
\displaystyle =\frac{\frac{\sin^2 A}{\cos^2 A}}{\frac{1}{\cos^2 A}}+\frac{\frac{\cos^2 A}{\sin^2 A}}{\frac{1}{\sin^2 A}}
\displaystyle =\sin^2 A+\cos^2 A
\displaystyle =1=\text{RHS}
\displaystyle \therefore \frac{\tan^2 A}{1+\tan^2 A}+\frac{\cot^2 A}{1+\cot^2 A}=1.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}=\cot\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}
\displaystyle =\frac{1+\cos\theta-(1-\cos^2\theta)}{\sin\theta(1+\cos\theta)}
\displaystyle \qquad[\because \sin^2\theta=1-\cos^2\theta]
\displaystyle =\frac{\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta)}
\displaystyle =\frac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}
\displaystyle =\frac{\cos\theta}{\sin\theta}
\displaystyle =\cot\theta=\text{RHS}
\displaystyle \therefore \frac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}=\cot\theta.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Prove the following trigonometric identity:}
\displaystyle \tan^2 A+\cot^2 A=\sec^2 A\,\mathrm{cosec}^2 A-2
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan^2 A+\cot^2 A
\displaystyle =\tan^2 A+\cot^2 A+2-2
\displaystyle =(\tan A+\cot A)^2-2
\displaystyle =\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)^2-2
\displaystyle =\left(\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}\right)^2-2
\displaystyle =\frac{1}{\sin^2 A\cos^2 A}-2\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle =\sec^2 A\,\mathrm{cosec}^2 A-2
\displaystyle =\text{RHS}
\displaystyle \therefore \tan^2 A+\cot^2 A=\sec^2 A\,\mathrm{cosec}^2 A-2.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\mathrm{cosec} A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}
\displaystyle =\frac{\tan A(1-\sec A)-\tan A(1+\sec A)}{(1+\sec A)(1-\sec A)}
\displaystyle =\frac{-2\tan A\sec A}{1-\sec^2 A}
\displaystyle =\frac{-2\tan A\sec A}{-\tan^2 A}\qquad[\because \sec^2 A-1=\tan^2 A]
\displaystyle =\frac{2\sec A}{\tan A}
\displaystyle =2\left(\frac{1/\cos A}{\sin A/\cos A}\right)
\displaystyle =\frac{2}{\sin A}
\displaystyle =2\mathrm{cosec} A=\text{RHS}
\displaystyle \therefore \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\mathrm{cosec} A.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }1+\frac{\cot^2\theta}{1+\mathrm{cosec}\theta}=\mathrm{cosec}\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=1+\frac{\cot^2\theta}{1+\mathrm{cosec}\theta}
\displaystyle =1+\frac{\mathrm{cosec}^2\theta-1}{1+\mathrm{cosec}\theta}
\displaystyle \qquad[\because \mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle =1+\frac{(\mathrm{cosec}\theta-1)(\mathrm{cosec}\theta+1)}{1+\mathrm{cosec}\theta}
\displaystyle =1+\mathrm{cosec}\theta-1
\displaystyle =\mathrm{cosec}\theta=\text{RHS}
\displaystyle \therefore 1+\frac{\cot^2\theta}{1+\mathrm{cosec}\theta}=\mathrm{cosec}\theta.

\displaystyle \text{(ii) }\frac{\cot A-1}{2-\sec^2 A}=\frac{\cot A}{1+\tan A}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cot A-1}{2-\sec^2 A}
\displaystyle =\frac{\frac{1}{\tan A}-1}{2-(1+\tan^2 A)}
\displaystyle \qquad[\because \cot A=\frac{1}{\tan A}\text{ and }\sec^2 A=1+\tan^2 A]
\displaystyle =\frac{\frac{1-\tan A}{\tan A}}{1-\tan^2 A}
\displaystyle =\frac{1-\tan A}{\tan A(1-\tan A)(1+\tan A)}
\displaystyle =\frac{1}{\tan A(1+\tan A)}
\displaystyle =\frac{\cot A}{1+\tan A}=\text{RHS}
\displaystyle \therefore \frac{\cot A-1}{2-\sec^2 A}=\frac{\cot A}{1+\tan A}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\cos\theta}{\mathrm{cosec}\theta+1}+\frac{\cos\theta}{\mathrm{cosec}\theta-1}=2\tan\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos\theta}{\mathrm{cosec}\theta+1}+\frac{\cos\theta}{\mathrm{cosec}\theta-1}
\displaystyle =\frac{\cos\theta}{\frac{1}{\sin\theta}+1}+\frac{\cos\theta}{\frac{1}{\sin\theta}-1}
\displaystyle =\frac{\sin\theta\cos\theta}{1+\sin\theta}+\frac{\sin\theta\cos\theta}{1-\sin\theta}
\displaystyle =\sin\theta\cos\theta\left(\frac{1-\sin\theta+1+\sin\theta}{1-\sin^2\theta}\right)
\displaystyle =\sin\theta\cos\theta\left(\frac{2}{\cos^2\theta}\right)\qquad[\because 1-\sin^2\theta=\cos^2\theta]
\displaystyle =\frac{2\sin\theta}{\cos\theta}
\displaystyle =2\tan\theta=\text{RHS}
\displaystyle \therefore \frac{\cos\theta}{\mathrm{cosec}\theta+1}+\frac{\cos\theta}{\mathrm{cosec}\theta-1}=2\tan\theta.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Prove the following trigonometric identity:}
\displaystyle \sec A(1-\sin A)(\sec A+\tan A)=1\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sec A(1-\sin A)(\sec A+\tan A)
\displaystyle =\frac{1}{\cos A}(1-\sin A)\left(\frac{1+\sin A}{\cos A}\right)
\displaystyle =\frac{(1-\sin A)(1+\sin A)}{\cos^2 A}
\displaystyle =\frac{1-\sin^2 A}{\cos^2 A}
\displaystyle =\frac{\cos^2 A}{\cos^2 A}\qquad[\because 1-\sin^2 A=\cos^2 A]
\displaystyle =1=\text{RHS}
\displaystyle \therefore \sec A(1-\sin A)(\sec A+\tan A)=1.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }(1+\tan^2 A)+\left(1+\frac{1}{\tan^2 A}\right)=\frac{1}{\sin^2 A-\sin^4 A}\hfill\text{[CBSE 2006C, 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\tan^2 A)+\left(1+\frac{1}{\tan^2 A}\right)
\displaystyle =\sec^2 A+(1+\cot^2 A)
\displaystyle =\sec^2 A+\mathrm{cosec}^2 A
\displaystyle =\frac{1}{\cos^2 A}+\frac{1}{\sin^2 A}
\displaystyle =\frac{\sin^2 A+\cos^2 A}{\sin^2 A\cos^2 A}
\displaystyle =\frac{1}{\sin^2 A\cos^2 A}\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle \text{RHS}=\frac{1}{\sin^2 A-\sin^4 A}
\displaystyle =\frac{1}{\sin^2 A(1-\sin^2 A)}
\displaystyle =\frac{1}{\sin^2 A\cos^2 A}\qquad[\because 1-\sin^2 A=\cos^2 A]
\displaystyle \therefore \text{LHS}=\text{RHS}.

\displaystyle \text{(ii) }\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\sin A}-\sin A\right)=\frac{1}{\tan A+\cot A}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\sin A}-\sin A\right)
\displaystyle =\left(\frac{1-\cos^2 A}{\cos A}\right)\left(\frac{1-\sin^2 A}{\sin A}\right)
\displaystyle =\frac{\sin^2 A}{\cos A}\times\frac{\cos^2 A}{\sin A}
\displaystyle =\sin A\cos A
\displaystyle \text{RHS}=\frac{1}{\tan A+\cot A}
\displaystyle =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}
\displaystyle =\frac{1}{\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}}
\displaystyle =\sin A\cos A\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Prove the following trigonometric identity:}
\displaystyle \cot^2 A\,\mathrm{cosec}^2 B-\cot^2 B\,\mathrm{cosec}^2 A=\cot^2 A-\cot^2 B
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cot^2 A\,\mathrm{cosec}^2 B-\cot^2 B\,\mathrm{cosec}^2 A
\displaystyle =\cot^2 A(1+\cot^2 B)-\cot^2 B(1+\cot^2 A)
\displaystyle \qquad[\because \mathrm{cosec}^2\theta=1+\cot^2\theta]
\displaystyle =\cot^2 A+\cot^2 A\cot^2 B-\cot^2 B-\cot^2 A\cot^2 B
\displaystyle =\cot^2 A-\cot^2 B
\displaystyle =\text{RHS}
\displaystyle \therefore \cot^2 A\,\mathrm{cosec}^2 B-\cot^2 B\,\mathrm{cosec}^2 A=\cot^2 A-\cot^2 B.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{(i) If }x=a\sec\theta+b\tan\theta\text{ and }y=a\tan\theta+b\sec\theta,
\displaystyle \text{prove that }x^2-y^2=a^2-b^2.\hfill\text{[CBSE 2001, 2002C, 2025]}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=(a\sec\theta+b\tan\theta)^2-(a\tan\theta+b\sec\theta)^2
\displaystyle =a^2\sec^2\theta+b^2\tan^2\theta+2ab\sec\theta\tan\theta
\displaystyle \quad-a^2\tan^2\theta-b^2\sec^2\theta-2ab\sec\theta\tan\theta
\displaystyle =a^2(\sec^2\theta-\tan^2\theta)-b^2(\sec^2\theta-\tan^2\theta)
\displaystyle =(a^2-b^2)(\sec^2\theta-\tan^2\theta)
\displaystyle =a^2-b^2\qquad[\because \sec^2\theta-\tan^2\theta=1]
\displaystyle \therefore x^2-y^2=a^2-b^2.

\displaystyle \text{(ii) If }x\cos\theta+y\sin\theta=a\text{ and }x\sin\theta-y\cos\theta=b,
\displaystyle \text{prove that }a^2+b^2=x^2+y^2.\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle a^2+b^2=(x\cos\theta+y\sin\theta)^2+(x\sin\theta-y\cos\theta)^2
\displaystyle =x^2\cos^2\theta+y^2\sin^2\theta+2xy\sin\theta\cos\theta
\displaystyle \quad+x^2\sin^2\theta+y^2\cos^2\theta-2xy\sin\theta\cos\theta
\displaystyle =x^2(\sin^2\theta+\cos^2\theta)+y^2(\sin^2\theta+\cos^2\theta)
\displaystyle =x^2+y^2\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle \therefore a^2+b^2=x^2+y^2.

\displaystyle \text{(iii) Use the identity }\sin^2 A+\cos^2 A=1\text{ to prove that }\tan^2 A+1=\sec^2 A.
\displaystyle \text{Hence, find the value of }\tan A,\text{ when }\sec A=\frac{5}{3},\text{ where }A\text{ is an acute angle.}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \sin^2 A+\cos^2 A=1
\displaystyle \text{Dividing throughout by }\cos^2 A,
\displaystyle \frac{\sin^2 A}{\cos^2 A}+\frac{\cos^2 A}{\cos^2 A}=\frac{1}{\cos^2 A}
\displaystyle \therefore \tan^2 A+1=\sec^2 A.
\displaystyle \text{Now, }\sec A=\frac{5}{3}
\displaystyle \tan^2 A=\sec^2 A-1
\displaystyle =\left(\frac{5}{3}\right)^2-1
\displaystyle =\frac{25}{9}-\frac{9}{9}=\frac{16}{9}
\displaystyle \therefore \tan A=\frac{4}{3}\qquad[\because A\text{ is an acute angle}]
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{(i) If }x=a\sec\theta+b\tan\theta\text{ and }y=a\tan\theta+b\sec\theta,
\displaystyle \text{prove that }x^2-y^2=a^2-b^2.\hfill\text{[CBSE 2001, 2002C]}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=(a\sec\theta+b\tan\theta)^2-(a\tan\theta+b\sec\theta)^2
\displaystyle =a^2\sec^2\theta+b^2\tan^2\theta+2ab\sec\theta\tan\theta
\displaystyle \quad-a^2\tan^2\theta-b^2\sec^2\theta-2ab\sec\theta\tan\theta
\displaystyle =a^2(\sec^2\theta-\tan^2\theta)-b^2(\sec^2\theta-\tan^2\theta)
\displaystyle =(a^2-b^2)(\sec^2\theta-\tan^2\theta)
\displaystyle =a^2-b^2\qquad[\because \sec^2\theta-\tan^2\theta=1]
\displaystyle \therefore x^2-y^2=a^2-b^2.

\displaystyle \text{(ii) If }x\cos\theta+y\sin\theta=a\text{ and }x\sin\theta-y\cos\theta=b,
\displaystyle \text{prove that }a^2+b^2=x^2+y^2.\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle a^2+b^2=(x\cos\theta+y\sin\theta)^2+(x\sin\theta-y\cos\theta)^2
\displaystyle =x^2\cos^2\theta+y^2\sin^2\theta+2xy\sin\theta\cos\theta
\displaystyle \quad+x^2\sin^2\theta+y^2\cos^2\theta-2xy\sin\theta\cos\theta
\displaystyle =x^2(\sin^2\theta+\cos^2\theta)+y^2(\sin^2\theta+\cos^2\theta)
\displaystyle =x^2+y^2\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle \therefore a^2+b^2=x^2+y^2.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }\mathrm{cosec}\theta+\cot\theta=m\text{ and }\mathrm{cosec}\theta-\cot\theta=n,
\displaystyle \text{prove that }mn=1.
\displaystyle \text{Answer:}
\displaystyle m=\mathrm{cosec}\theta+\cot\theta
\displaystyle n=\mathrm{cosec}\theta-\cot\theta
\displaystyle \therefore mn=(\mathrm{cosec}\theta+\cot\theta)(\mathrm{cosec}\theta-\cot\theta)
\displaystyle =\mathrm{cosec}^2\theta-\cot^2\theta
\displaystyle =1\qquad[\because \mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle \therefore mn=1.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Prove the following trigonometric identities:}
\displaystyle \text{(i) }\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}
\displaystyle =\frac{(1+\cos\theta+\sin\theta)(1-\cos\theta+\sin\theta)}{(1+\cos\theta-\sin\theta)(1-\cos\theta+\sin\theta)}
\displaystyle =\frac{(1+\sin\theta)^2-\cos^2\theta}{1-(\cos\theta-\sin\theta)^2}
\displaystyle =\frac{2\sin\theta(1+\sin\theta)}{2\sin\theta\cos\theta}
\displaystyle \qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle =\frac{1+\sin\theta}{\cos\theta}=\text{RHS}
\displaystyle \therefore \frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}.

\displaystyle \text{(ii) }\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1}{\sec\theta-\tan\theta}\hfill\text{[CBSE 2001, 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}
\displaystyle =\frac{(1+\sin\theta-\cos\theta)(1+\sin\theta+\cos\theta)}{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}
\displaystyle =\frac{(1+\sin\theta)^2-\cos^2\theta}{(\sin\theta+\cos\theta)^2-1}
\displaystyle =\frac{2\sin\theta(1+\sin\theta)}{2\sin\theta\cos\theta}
\displaystyle =\frac{1+\sin\theta}{\cos\theta}
\displaystyle =\sec\theta+\tan\theta
\displaystyle =\frac{1}{\sec\theta-\tan\theta}\qquad[\because (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1]
\displaystyle =\text{RHS}
\displaystyle \therefore \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1}{\sec\theta-\tan\theta}.

\displaystyle \text{(iii) }\frac{\cos\theta-\sin\theta+1}{\cos\theta+\sin\theta-1}=\mathrm{cosec}\theta+\cot\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos\theta-\sin\theta+1}{\cos\theta+\sin\theta-1}
\displaystyle =\frac{(1+\cos\theta-\sin\theta)(1+\cos\theta+\sin\theta)}{(\cos\theta+\sin\theta-1)(\cos\theta+\sin\theta+1)}
\displaystyle =\frac{(1+\cos\theta)^2-\sin^2\theta}{(\cos\theta+\sin\theta)^2-1}
\displaystyle =\frac{2\cos\theta(1+\cos\theta)}{2\sin\theta\cos\theta}
\displaystyle =\frac{1+\cos\theta}{\sin\theta}
\displaystyle =\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}
\displaystyle =\mathrm{cosec}\theta+\cot\theta=\text{RHS}
\displaystyle \therefore \frac{\cos\theta-\sin\theta+1}{\cos\theta+\sin\theta-1}=\mathrm{cosec}\theta+\cot\theta.

\displaystyle \text{(iv) }(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)=\sec\theta+\mathrm{cosec}\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)
\displaystyle =(\sin\theta+\cos\theta)\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)
\displaystyle =(\sin\theta+\cos\theta)\left(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right)
\displaystyle =\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle =\frac{1}{\cos\theta}+\frac{1}{\sin\theta}
\displaystyle =\sec\theta+\mathrm{cosec}\theta=\text{RHS}
\displaystyle \therefore (\sin\theta+\cos\theta)(\tan\theta+\cot\theta)=\sec\theta+\mathrm{cosec}\theta.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\frac{1-\sin\theta}{\cos\theta}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}
\displaystyle \text{Since }(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1,
\displaystyle \sec\theta+\tan\theta=\frac{1}{\sec\theta-\tan\theta}
\displaystyle \therefore \text{LHS}=\frac{1+\sec\theta-\tan\theta}{1+\frac{1}{\sec\theta-\tan\theta}}
\displaystyle =\sec\theta-\tan\theta
\displaystyle =\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}
\displaystyle =\frac{1-\sin\theta}{\cos\theta}=\text{RHS}
\displaystyle \therefore \frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\frac{1-\sin\theta}{\cos\theta}.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Prove the following trigonometric identity:}
\displaystyle (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\mathrm{cosec}^2 A}-\frac{\mathrm{cosec} A}{\sec^2 A}
\displaystyle =\sin A\tan A-\cot A\cos A.\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\cot A+\tan A)(\sin A-\cos A)
\displaystyle =\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)(\sin A-\cos A)
\displaystyle =\left(\frac{\sin A\cos A+\cos^2 A+\sin^2 A}{\sin A\cos A}\right)(\sin A-\cos A)
\displaystyle =\frac{(1+\sin A\cos A)(\sin A-\cos A)}{\sin A\cos A}
\displaystyle \qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle =\frac{(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)}{\sin A\cos A}
\displaystyle =\frac{\sin^3 A-\cos^3 A}{\sin A\cos A}
\displaystyle =\frac{\sin^2 A}{\cos A}-\frac{\cos^2 A}{\sin A}
\displaystyle =\sin A\tan A-\cot A\cos A
\displaystyle \text{Also, }\frac{\sec A}{\mathrm{cosec}^2 A}-\frac{\mathrm{cosec} A}{\sec^2 A}
\displaystyle =\frac{1/\cos A}{1/\sin^2 A}-\frac{1/\sin A}{1/\cos^2 A}
\displaystyle =\frac{\sin^2 A}{\cos A}-\frac{\cos^2 A}{\sin A}
\displaystyle =\sin A\tan A-\cot A\cos A
\displaystyle \therefore (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\mathrm{cosec}^2 A}-\frac{\mathrm{cosec} A}{\sec^2 A}
\displaystyle =\sin A\tan A-\cot A\cos A.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\sec\theta\,\mathrm{cosec}\theta-2\sin\theta\cos\theta
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}
\displaystyle =\frac{\tan^3\theta}{\sec^2\theta}+\frac{\cot^3\theta}{\mathrm{cosec}^2\theta}
\displaystyle =\frac{\sin^3\theta}{\cos\theta}+\frac{\cos^3\theta}{\sin\theta}
\displaystyle =\frac{\sin^4\theta+\cos^4\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{1-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{1}{\sin\theta\cos\theta}-2\sin\theta\cos\theta
\displaystyle =\sec\theta\,\mathrm{cosec}\theta-2\sin\theta\cos\theta=\text{RHS}
\displaystyle \therefore \frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}=\sec\theta\,\mathrm{cosec}\theta-2\sin\theta\cos\theta.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Prove the following trigonometric identity:}
\displaystyle \frac{\tan A}{(1+\tan^2 A)^2}+\frac{\cot A}{(1+\cot^2 A)^2}=\sin A\cos A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\tan A}{(1+\tan^2 A)^2}+\frac{\cot A}{(1+\cot^2 A)^2}
\displaystyle =\frac{\tan A}{\sec^4 A}+\frac{\cot A}{\mathrm{cosec}^4 A}
\displaystyle =\frac{\sin A}{\cos A}\cos^4 A+\frac{\cos A}{\sin A}\sin^4 A
\displaystyle =\sin A\cos^3 A+\cos A\sin^3 A
\displaystyle =\sin A\cos A(\cos^2 A+\sin^2 A)
\displaystyle =\sin A\cos A\qquad[\because \sin^2 A+\cos^2 A=1]
\displaystyle =\text{RHS}
\displaystyle \therefore \frac{\tan A}{(1+\tan^2 A)^2}+\frac{\cot A}{(1+\cot^2 A)^2}=\sin A\cos A.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Prove the following trigonometric identity:}
\displaystyle \left(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2=\frac{1-\cos\theta}{1+\cos\theta}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\left(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2
\displaystyle \text{Now, }\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}
\displaystyle \text{since }\sin\theta(1+\sin\theta-\cos\theta)=(1-\cos\theta)(1+\sin\theta+\cos\theta)
\displaystyle \text{as both sides simplify to }\sin\theta+\sin^2\theta-\sin\theta\cos\theta.
\displaystyle \therefore \text{LHS}=\left(\frac{1-\cos\theta}{\sin\theta}\right)^2
\displaystyle =\frac{(1-\cos\theta)^2}{\sin^2\theta}
\displaystyle =\frac{(1-\cos\theta)^2}{1-\cos^2\theta}
\displaystyle =\frac{1-\cos\theta}{1+\cos\theta}=\text{RHS}
\displaystyle \therefore \left(\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2=\frac{1-\cos\theta}{1+\cos\theta}.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }3\sin\theta+5\cos\theta=5,\text{ prove that }5\sin\theta-3\cos\theta=\pm3.
\displaystyle \text{Answer:}
\displaystyle (3\sin\theta+5\cos\theta)^2+(5\sin\theta-3\cos\theta)^2
\displaystyle =9\sin^2\theta+25\cos^2\theta+30\sin\theta\cos\theta
\displaystyle \quad+25\sin^2\theta+9\cos^2\theta-30\sin\theta\cos\theta
\displaystyle =34(\sin^2\theta+\cos^2\theta)
\displaystyle =34
\displaystyle \text{Given, }3\sin\theta+5\cos\theta=5.
\displaystyle \therefore 5^2+(5\sin\theta-3\cos\theta)^2=34
\displaystyle (5\sin\theta-3\cos\theta)^2=34-25=9
\displaystyle \therefore 5\sin\theta-3\cos\theta=\pm3.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }\sin\theta+2\cos\theta=1,\text{ find the value of }2\sin\theta-\cos\theta.
\displaystyle \text{Answer:}
\displaystyle (2\sin\theta-\cos\theta)^2+(\sin\theta+2\cos\theta)^2
\displaystyle =4\sin^2\theta+\cos^2\theta-4\sin\theta\cos\theta
\displaystyle \quad+\sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta
\displaystyle =5(\sin^2\theta+\cos^2\theta)
\displaystyle =5
\displaystyle \text{Given, }\sin\theta+2\cos\theta=1.
\displaystyle \therefore (2\sin\theta-\cos\theta)^2+1^2=5
\displaystyle (2\sin\theta-\cos\theta)^2=4
\displaystyle \therefore 2\sin\theta-\cos\theta=\pm2.
\displaystyle \text{Thus, the given information alone does not establish the value as }2\text{ only.}
\displaystyle \text{An additional condition is required to select the positive value }2.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If }T_n=\sin^n\theta+\cos^n\theta,\text{ prove that }\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}.
\displaystyle \text{Answer:}
\displaystyle T_3-T_5=(\sin^3\theta+\cos^3\theta)-(\sin^5\theta+\cos^5\theta)
\displaystyle =\sin^3\theta(1-\sin^2\theta)+\cos^3\theta(1-\cos^2\theta)
\displaystyle =\sin^3\theta\cos^2\theta+\cos^3\theta\sin^2\theta
\displaystyle =\sin^2\theta\cos^2\theta(\sin\theta+\cos\theta)
\displaystyle =\sin^2\theta\cos^2\theta\,T_1
\displaystyle \therefore \frac{T_3-T_5}{T_1}=\sin^2\theta\cos^2\theta.
\displaystyle \text{Also, }T_5-T_7=(\sin^5\theta+\cos^5\theta)-(\sin^7\theta+\cos^7\theta)
\displaystyle =\sin^5\theta(1-\sin^2\theta)+\cos^5\theta(1-\cos^2\theta)
\displaystyle =\sin^5\theta\cos^2\theta+\cos^5\theta\sin^2\theta
\displaystyle =\sin^2\theta\cos^2\theta(\sin^3\theta+\cos^3\theta)
\displaystyle =\sin^2\theta\cos^2\theta\,T_3
\displaystyle \therefore \frac{T_5-T_7}{T_3}=\sin^2\theta\cos^2\theta.
\displaystyle \therefore \frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{If }a\cos^3\theta+3a\cos\theta\sin^2\theta=m,
\displaystyle a\sin^3\theta+3a\cos^2\theta\sin\theta=n,\text{ prove that}
\displaystyle (m+n)^{2/3}+(m-n)^{2/3}=2a^{2/3}.
\displaystyle \text{Answer:}
\displaystyle m+n=a\cos^3\theta+a\sin^3\theta+3a\cos\theta\sin^2\theta+3a\cos^2\theta\sin\theta
\displaystyle =a[\cos^3\theta+\sin^3\theta+3\sin\theta\cos\theta(\sin\theta+\cos\theta)]
\displaystyle =a(\sin\theta+\cos\theta)^3
\displaystyle \therefore (m+n)^{2/3}=a^{2/3}(\sin\theta+\cos\theta)^2.
\displaystyle \text{Also, }m-n=a\cos^3\theta-a\sin^3\theta+3a\cos\theta\sin^2\theta-3a\cos^2\theta\sin\theta
\displaystyle =a[\cos^3\theta-\sin^3\theta-3\sin\theta\cos\theta(\cos\theta-\sin\theta)]
\displaystyle =a(\cos\theta-\sin\theta)^3
\displaystyle \therefore (m-n)^{2/3}=a^{2/3}(\cos\theta-\sin\theta)^2.
\displaystyle \therefore (m+n)^{2/3}+(m-n)^{2/3}
\displaystyle =a^{2/3}[(\sin\theta+\cos\theta)^2+(\cos\theta-\sin\theta)^2]
\displaystyle =a^{2/3}[2\sin^2\theta+2\cos^2\theta]
\displaystyle =2a^{2/3}(\sin^2\theta+\cos^2\theta)
\displaystyle =2a^{2/3}.
\displaystyle \therefore (m+n)^{2/3}+(m-n)^{2/3}=2a^{2/3}.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{If }x=a\cos^3\theta,\ y=b\sin^3\theta,\text{ prove that}
\displaystyle \left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1.
\displaystyle \text{Answer:}
\displaystyle x=a\cos^3\theta
\displaystyle \therefore \frac{x}{a}=\cos^3\theta
\displaystyle \therefore \left(\frac{x}{a}\right)^{2/3}=\cos^2\theta
\displaystyle \text{Also, }y=b\sin^3\theta
\displaystyle \therefore \frac{y}{b}=\sin^3\theta
\displaystyle \therefore \left(\frac{y}{b}\right)^{2/3}=\sin^2\theta
\displaystyle \therefore \left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}
\displaystyle =\cos^2\theta+\sin^2\theta
\displaystyle =1
\displaystyle \therefore \left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{If }a\cos\theta+b\sin\theta=m\text{ and }a\sin\theta-b\cos\theta=n,
\displaystyle \text{prove that }a^2+b^2=m^2+n^2.
\displaystyle \text{Answer:}
\displaystyle m^2+n^2=(a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2
\displaystyle =a^2\cos^2\theta+b^2\sin^2\theta+2ab\sin\theta\cos\theta
\displaystyle \quad+a^2\sin^2\theta+b^2\cos^2\theta-2ab\sin\theta\cos\theta
\displaystyle =a^2(\sin^2\theta+\cos^2\theta)+b^2(\sin^2\theta+\cos^2\theta)
\displaystyle =a^2+b^2\qquad[\because \sin^2\theta+\cos^2\theta=1]
\displaystyle \therefore a^2+b^2=m^2+n^2.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }\cos A+\cos^2 A=1,\text{ prove that }\sin^2 A+\sin^4 A=1.
\displaystyle \hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos A+\cos^2 A=1
\displaystyle \therefore \cos A=1-\cos^2 A
\displaystyle \therefore \cos A=\sin^2 A\qquad[\because 1-\cos^2 A=\sin^2 A]
\displaystyle \therefore \sin^4 A=\cos^2 A
\displaystyle \therefore \sin^2 A+\sin^4 A=\cos A+\cos^2 A
\displaystyle =1
\displaystyle \therefore \sin^2 A+\sin^4 A=1.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If }\cos\theta+\cos^2\theta=1,\text{ prove that}
\displaystyle \sin^{12}\theta+3\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos\theta+\cos^2\theta=1
\displaystyle \therefore \cos\theta=1-\cos^2\theta=\sin^2\theta
\displaystyle \text{Let }x=\sin^2\theta.
\displaystyle \therefore x=\cos\theta
\displaystyle \therefore x+x^2=1
\displaystyle \text{LHS}=x^6+3x^5+3x^4+x^3+2x^2+2x-2
\displaystyle =x^3(x^3+3x^2+3x+1)+2(x^2+x-1)
\displaystyle =x^3(x+1)^3+2(x^2+x-1)
\displaystyle =[x(x+1)]^3+2(x^2+x-1)
\displaystyle =(x^2+x)^3+2(x^2+x-1)
\displaystyle =1^3+2(1-1)
\displaystyle =1=\text{RHS}
\displaystyle \therefore \sin^{12}\theta+3\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Given that }(1+\cos\alpha)(1+\cos\beta)(1+\cos\gamma)
\displaystyle =(1-\cos\alpha)(1-\cos\beta)(1-\cos\gamma),\text{ show that one of the values of each}
\displaystyle \text{member of this equality is }\sin\alpha\sin\beta\sin\gamma.
\displaystyle \text{Answer:}
\displaystyle \text{Let }(1+\cos\alpha)(1+\cos\beta)(1+\cos\gamma)
\displaystyle =(1-\cos\alpha)(1-\cos\beta)(1-\cos\gamma)=k.
\displaystyle \therefore k^2=(1+\cos\alpha)(1-\cos\alpha)(1+\cos\beta)(1-\cos\beta)
\displaystyle \qquad\times(1+\cos\gamma)(1-\cos\gamma)
\displaystyle =(1-\cos^2\alpha)(1-\cos^2\beta)(1-\cos^2\gamma)
\displaystyle =\sin^2\alpha\sin^2\beta\sin^2\gamma
\displaystyle \therefore k=\sin\alpha\sin\beta\sin\gamma\qquad[\because \alpha,\beta,\gamma\text{ are acute angles}]
\displaystyle \therefore \text{Each member of the given equality has the value }\sin\alpha\sin\beta\sin\gamma.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{If }x=a\sec\theta\cos\phi,\ y=b\sec\theta\sin\phi\text{ and }z=c\tan\theta,
\displaystyle \text{show that }\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1.
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}
\displaystyle =\frac{a^2\sec^2\theta\cos^2\phi}{a^2}+\frac{b^2\sec^2\theta\sin^2\phi}{b^2}-\frac{c^2\tan^2\theta}{c^2}
\displaystyle =\sec^2\theta\cos^2\phi+\sec^2\theta\sin^2\phi-\tan^2\theta
\displaystyle =\sec^2\theta(\cos^2\phi+\sin^2\phi)-\tan^2\theta
\displaystyle =\sec^2\theta-\tan^2\theta\qquad[\because \sin^2\phi+\cos^2\phi=1]
\displaystyle =1
\displaystyle \therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1.
\displaystyle \\

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