Question

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

• A. $-\frac{1}{2}$
• B. 2
• C. $\sin x+\cos x$
• D. $\sin x-\cos x$

Answer 1

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

Solution:

$y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

$={\cot }^{-1}\left[\frac{\sqrt{{\sin }^{2}x{\cos }^{2}x+2\sin \frac{x}{2}\cos \frac{x}{2}}+\sqrt{{\sin }^2+{\cos }^{2}x-2\sin \frac{x}{2}\cos \frac{x}{2}}}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin \frac{x}{2}\cos \frac{x}{2}}-\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin \frac{x}{2}}\cos \frac{x}{2}}\right]$

$={\cot }^{-1}\left[\frac{|\sin \frac{x}{2}+\cos \frac{x}{2}|+|\sin \frac{x}{2}-\cos \frac{x}{2}|}{|\sin \frac{x}{2}+\cos \frac{x}{2}|-|\sin \frac{x}{2}-\cos \frac{x}{2}|}\right]$

As $0<x<\frac{\pi }{2}$

$\therefore y={\cot }^{-1}\left[\frac{(\sin \frac{x}{2}+\cos \frac{x}{2})+(\sin \frac{x}{2}-\cos \frac{x}{2})}{(\sin \frac{x}{2}+\cos \frac{x}{2})-(\sin \frac{x}{2}-\cos \frac{x}{2})}\right]$

$={\cot }^{-1}\left(\tan \frac{x}{2}\right)$

$={\cot }^{-1}\left[\cot (\frac{\pi }{2}-\frac{x}{2})\right]$

$=\frac{\pi }{2}-\frac{x}{2}$

$\therefore \frac{dy}{dx}=-\frac{1}{2}$