Choose the correct alternative in the following:

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

$\Rightarrow \frac{d}{d x}\left\{\tan ^{-1}\left(\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right)\right\} \because \sin \left(\frac{\pi}{2}-x\right)=\cos x$ and $\cos \left(\frac{\pi}{2}-x\right)$

$=\sin x$

Put $\frac{\pi}{2}-x=2 t \Rightarrow t=\frac{\pi}{4}-\frac{x}{2}-(1)$

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

$\Rightarrow \frac{d}{d x}\left\{\tan ^{-1}\left(\frac{2 \sin t \cdot \cos t}{2 \cos ^{2} t}\right)\right\} \because \sin 2 t=2 \sin t \cdot \cos t$ and $1+\cos 2 t=2 \cos ^{2} t$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left\{\tan ^{-1}(\tan \mathrm{t})\right\}$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{t}\} \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}-\frac{x}{2}\right)=-\frac{1}{2}-$ From $(1)$

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