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If $y=\tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$ is
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The correct answer is:
$-\frac{1}{2}$
$$
\text { Hints : } \begin{aligned}
& y=\tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \\
& =\tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}=\tan ^{-1}\left|\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|=\left(\frac{\pi}{4}-\frac{x}{2}\right) \\
& \frac{d y}{d x}=-\frac{1}{2}
\end{aligned}
$$
\text { Hints : } \begin{aligned}
& y=\tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \\
& =\tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}=\tan ^{-1}\left|\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|=\left(\frac{\pi}{4}-\frac{x}{2}\right) \\
& \frac{d y}{d x}=-\frac{1}{2}
\end{aligned}
$$
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