Question

If $y=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right), 0<x<\frac{\pi}{2}$, then $\frac{d y}{d x}$ is equal to

(A) $\frac{1}{2}$

(B) 2

(C) -2

(D) $-\frac{1}{2}$

Answer 1

HINT/SOLUTION —

$\begin{aligned}y & =\tan ^{-1}\left(\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}\right) \\& =\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \\& =\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\end{aligned}$

So, The correct option is (D).