#### Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 15

$-\frac{1}{2}$

Hint:

Differentiate the function w.r.t x

Given:

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

Solution:

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

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

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

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

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

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

\begin{aligned} &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=0-\frac{1}{2} \\\\ &\frac{d u}{d x}=-\frac{1}{2} \end{aligned}