Question

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

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

Answer 1

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

Simplyfying,
$\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}=\frac{2+2\cos x}{2\sin x}=\frac{1+\cos x}{\sin x}=\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\cot \frac{x}{2}$
So,
$y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]\Rightarrow y={\cot }^{-1}\left[\cot \frac{x}{2}\right]$
As
$0<x<\pi /2\Rightarrow 0<\frac{x}{2}<\frac{\pi }{4}$
$\therefore y=\frac{x}{2}\Rightarrow \frac{dy}{dx}=\frac{1}{2}$