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If $y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{-1}{4}$
Given $y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]$
$$
=\tan ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=\tan ^{-1}\left(\cot \frac{x}{4}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right]
$$
$\therefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}$
$$
=\tan ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=\tan ^{-1}\left(\cot \frac{x}{4}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right]
$$
$\therefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}$
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