Question

If $y={\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right),$ $0<x<1$ then $\frac{dy}{dx}$ equals-

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

Answer 1

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

$y={\tan }^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right]$
Put $x=\cos \theta$
$\Rightarrow y={\tan }^{-1}\left[\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\right]$
$\Rightarrow y={\tan }^{-1}\left[\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\right]$
$\Rightarrow y={\tan }^{-1}\left[\tan (\frac{\pi }{4}+\frac{\theta }{2})\right]$
$\Rightarrow y=\frac{\pi }{4}+\frac{\theta }{2}$
$\Rightarrow y=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^2}}$