If y = tan^{-1} left[ sqrt{frac{1 + cos(x/2)} | vedclass

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(B) Given $y = 	an^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} ight]$.Using the trigonometric identities $1 + \cos 	heta = 2 \cos^2(	h

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$\frac{-1}{4}$

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(B) Given $y = 	an^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} ight]$.Using the trigonometric identities $1 + \cos 	heta = 2 \cos^2(	heta/2)$ and $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we have:$y = 	an^{-1} \sqrt{\frac{2 \cos^2(x/4)}{2 \sin^2(x/4)}}$$y = 	an^{-1} \sqrt{\cot^2(x/4)}$$y = 	an^{-1} (\cot(x/4))$Since $\cot 	heta = 	an(\pi/2 - 	heta)$,we get:$y = 	an^{-1} \left[ 	an \left( \frac{\pi}{2} - \frac{x}{4} ight) ight]$$y = \frac{\pi}{2} - \frac{x}{4}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \frac{x}{4} ight) = 0 - \frac{1}{4} = -\frac{1}{4}$.

If $y = \tan^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right]$,then $\frac{dy}{dx} = $

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