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If $y=\cot ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{1}{2}$
$\begin{aligned}
y &=\cot ^{-1} \sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}}=\cot ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \\
&=\cot ^{-1}\left[\frac{1-\tan \frac{x}{2}}{\left.1+\tan \frac{x}{2}\right)}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\right]\right.\\
\therefore \frac{d y}{d x} &=0+\frac{1}{2}=\frac{1}{2}
\end{aligned}$
y &=\cot ^{-1} \sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}}=\cot ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \\
&=\cot ^{-1}\left[\frac{1-\tan \frac{x}{2}}{\left.1+\tan \frac{x}{2}\right)}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\right]\right.\\
\therefore \frac{d y}{d x} &=0+\frac{1}{2}=\frac{1}{2}
\end{aligned}$
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