# Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

If $f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \pi / 2$, then $f^{\prime}(\pi / 6)$ is

A. $-1 / 4$

B. $-1 / 2$

C. $1 / 4$

D. $1 / 2$

Solution:

$f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}$

$=\tan ^{-1} \sqrt{\frac{1+2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}{1-2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}}$

$\because \sin 2 x=2 \sin x \cos x$

$\Rightarrow \sin x=2 \sin x / 2 \cos x / 2$

$=\tan ^{-1} \sqrt{\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \cdot \sin \frac{x}{2} \cos \frac{x}{2}}}$

$\because \sin ^{2} x / 2+\cos ^{2} x / 2=1$

$=\tan ^{-1} \sqrt{\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}}}$

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

Dividing by $\cos x / 2$ we get

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

$=-\tan ^{-1}\left(\frac{\tan \frac{x}{2}+\tan \frac{\pi}{4}}{1-\tan \frac{x}{2} \cdot \tan \frac{\pi}{4}}\right)$

$=-\tan ^{-1}\left[\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right]$

$\because \tan (A-B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$

$=\tan ^{-1}\left[\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right] \because 0 \leq x \leq \pi / 2$

$\therefore f(x)=\left(\frac{x}{2}+\frac{\pi}{4}\right)$

$\mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{2}=(\mathrm{D})$