Question

If $y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right](0<x<\frac{\pi }{2})$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $3$
• D. $1$

Answer 1

The correct option is A $\frac{1}{2}$

$y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]={\cot }^{-1}[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}]={\cot }^{-1}\left[\frac{2+2\cos x}{2\sin x}\right]={\cot }^{-1}\left[\frac{1+\cos x}{\sin x}\right]={\cot }^{-1}\left[\cot \frac{x}{2}\right]=\frac{x}{2},x\in (0,\frac{\pi }{2})\therefore \frac{dy}{dx}=\frac{1}{2}$