Find the value of frac{d}{dx} left[ tan^{-1} | vedclass

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

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$\frac{1}{2}$

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(C) Given expression is $y = 	an^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.Using trigonometric identities,$1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting these,we get $y = 	an^{-1} \sqrt{\frac{2 \sin^2 (x/2)}{2 \cos^2 (x/2)}} = 	an^{-1} \sqrt{	an^2 \frac{x}{2}} = 	an^{-1} \left( 	an \frac{x}{2} ight)$.Thus,$y = \frac{x}{2}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} ight) = \frac{1}{2}$.

Find the value of $\frac{d}{dx} \left[ \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right]$.

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