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

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

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$\frac{1}{2}, 0$

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(C) Given $y = 	an^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.Using the trigonometric identities $1 - \cos x = 2 \sin^2(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$,we get:$y = 	an^{-1} \sqrt{\frac{2 \sin^2(x/2)}{2 \cos^2(x/2)}}$$y = 	an^{-1} \sqrt{	an^2(x/2)}$$y = 	an^{-1}(	an(x/2)) = \frac{x}{2}$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.Again,differentiating with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{2}) = 0$.Thus,the values are $\frac{1}{2}$ and $0$.

If $y = \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$,then the values of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ respectively are

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