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

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

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$\frac{-1}{2}$

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(B) Given $y = 	an^{-1} \sqrt{\frac{1 + \cos x}{1 - \cos x}}$.Using 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}}}$$y = 	an^{-1} \sqrt{\cot^2 \frac{x}{2}}$$y = 	an^{-1} \left( \cot \frac{x}{2} ight)$Since $\cot \frac{x}{2} = 	an \left( \frac{\pi}{2} - \frac{x}{2} ight)$,we have:$y = 	an^{-1} \left( 	an \left( \frac{\pi}{2} - \frac{x}{2} ight) ight)$$y = \frac{\pi}{2} - \frac{x}{2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \frac{x}{2} ight) = 0 - \frac{1}{2} = -\frac{1}{2}$.

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

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