Question:
$\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}
Solution:
Let $y=\tan ^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right]$
$=\tan ^{-1}\left[\sqrt{\frac{2 \sin ^{2} x / 2}{2 \cos ^{2} x / 2}}\right]\left[\begin{array}{r}\because 1-\cos x=2 \sin ^{2} x / 2 \\ 1+\cos x=2 \cos ^{2} x / 2\end{array}\right]$
$=\tan ^{-1}\left[\frac{\sin x / 2}{\cos x / 2}\right]=\tan ^{-1}\left[\tan \frac{x}{2}\right]$
Thus, $y=\frac{x}{2}$
Differentiating both sides w.r.t. $x$
$\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)=\frac{1}{2} \cdot 1=\frac{1}{2}$
Hence, $\frac{d y}{d x}=\frac{1}{2}$