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, $\frac{dy}{dx}=\pm \frac{1}{2}$