Question

The derivative of $y=\text{Tan}^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$ with respect to $x$ is equal to

• A. $-1$
• B. $0$
• C. $\pm 2$
• D. $\pm \frac{1}{2}$

Answer 1

Answer: (D)

It is given that,
$y=\tan ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$
$\because \sqrt{1+\sin x}=\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$=\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|$
and
$\sqrt{1-\sin x}=\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$=\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|$
$\therefore y=\tan ^{-1}\left(\cot \frac{x}{2}\right)$
when $\cos \frac{x}{2}+\sin \frac{x}{2}$ and
$\cos \frac{x}{2}-\sin \frac{x}{2}=\frac{\pi}{2}-\frac{x}{2}$ are positive
$\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}$,
when $\cos \frac{x}{2}+\sin \frac{x}{2}$ is positive
and $\cos \frac{x}{2}-\sin \frac{x}{2}$ is negative
$\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}$, when $\cos \frac{x}{2}+\sin \frac{x}{2}$ is negative
and $\cos \frac{x}{2}-\sin \frac{x}{2}$ is positive
$\tan ^{-1}\left(\cot \frac{x}{2}\right)=\frac{\pi}{2}-\frac{x}{2}$,
when $\cos \frac{x}{2}+\sin \frac{x}{2}$ and as
$\cos \frac{x}{2}-\sin \frac{x}{2}$ is negative.
So, $\frac{d y}{d x}=\pm \frac{1}{2}$