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If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{d y}{d x}=$
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$\frac{1}{2}$
$\begin{aligned} & y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}} \\ & =\tan ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}=\tan ^{-1}\left(\cot \frac{x}{2}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]=\frac{\pi}{2}-\frac{x}{2} \\ & \therefore \frac{d y}{d x}=0-\frac{1}{2}=\frac{-1}{2}\end{aligned}$
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