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

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(B) 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 =

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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 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(x/2)}{2\sin^2(x/2)}}$$y = 	an^{-1}\sqrt{\cot^2(x/2)}$$y = 	an^{-1}(\cot(x/2))$Since $\cot(x/2) = 	an(\frac{\pi}{2} - \frac{x}{2})$,we have:$y = 	an^{-1}(	an(\frac{\pi}{2} - \frac{x}{2}))$$y = \frac{\pi}{2} - \frac{x}{2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) - \frac{d}{dx}(\frac{x}{2})$$\frac{dy}{dx} = 0 - \frac{1}{2} = -\frac{1}{2}$.

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

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