Differentiate the following functions with respect to $x$ :
$\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right),-\pi
$y=\tan ^{-1}\left\{\frac{\sin x}{1+\cos x}\right\}$
Function $y$ is defined for all real numbers where $\cos x \neq-1$
Using $2 \cos ^{2} \theta=1+\cos 2 \theta$ and $2 \sin \theta \cos \theta=\sin 2 \theta$
$y=\tan ^{-1}\left\{\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right\}$
$y=\tan ^{-1}\left\{\tan \frac{x}{2}\right\}$
$y=\frac{x}{2}$
Differentiating w.r.t $x$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}$
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