Solve that equation
Question:

$\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}$

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, $\frac{d y}{d x}=\frac{1}{2}$

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