Question

$\frac{d}{dx}{\tan }^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=$

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

Answer 1

Answer: (C)

$\frac{1}{2}$

$\frac{d}{dx}{\tan }^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

$=\frac{d}{dx}{\tan }^{-1}\left[\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2})-(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)-(\sin \frac{x}{2}-\cos \frac{x}{2})}\right]$

$=\frac{d}{dx}{\tan }^{-1}\left[\tan \frac{x}{2}\right]$

$=\frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2}$

Hence, option C is correct.