If $y=\cot ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

$\begin{aligned} y &=\cot ^{-1} \sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}ight)^{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}ight)^{2}}}=\cot ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}ight) \ &=\cot ^{-1}\left[\frac{1-	an \frac{x}{2}}{\left.1+	an \frac{x}{2}ight)}=	an ^{-1}\left(\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}}ight)=	an ^{-1}\left[\frac{	an \frac{\pi}{4}+	an \frac{x}{2}}{1-	an \frac{\pi}{4} 	an \frac{x}{2}}ight]ight.\ 	herefore \frac{d y}{d x} &=0+\frac{1}{2}=\frac{1}{2} \end{aligned}$

If $y=\cot ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$, then $\frac{d y}{d x}=$

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