If y=tan ^{-1} sqrt{frac{1+cos x}{1-cos x}},t | vedclass

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(D) Given $y=	an ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$.Using the trigonometric identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $1-\cos x = 2 \sin^

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$\frac{-1}{2}$

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(D) Given $y=	an ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$.Using the trigonometric identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $1-\cos x = 2 \sin^2 \frac{x}{2}$,we get:$y=	an ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}$$y=	an ^{-1} \sqrt{\cot^2 \frac{x}{2}} = 	an ^{-1} \left(\cot \frac{x}{2}ight)$Since $\cot 	heta = 	an \left(\frac{\pi}{2} - 	hetaight)$,we have:$y=	an ^{-1} \left[	an \left(\frac{\pi}{2}-\frac{x}{2}ight)ight] = \frac{\pi}{2}-\frac{x}{2}$Differentiating with respect to $x$:$\frac{d y}{d x} = \frac{d}{d x} \left(\frac{\pi}{2} - \frac{x}{2}ight) = 0 - \frac{1}{2} = -\frac{1}{2}$

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$,then $\frac{d y}{d x}=$

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