If f(x) = tan^{-1}left( frac{sin x}{1 + cos x | vedclass

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(B) Given $f(x) = 	an^{-1}\left( \frac{\sin x}{1 + \cos x} ight)$.Using trigonometric identities,$\sin x = 2 \sin\left( \frac{x}{2} ight) \cos\le

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$\frac{1}{2}$

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(B) Given $f(x) = 	an^{-1}\left( \frac{\sin x}{1 + \cos x} ight)$.Using trigonometric identities,$\sin x = 2 \sin\left( \frac{x}{2} ight) \cos\left( \frac{x}{2} ight)$ and $1 + \cos x = 2 \cos^2\left( \frac{x}{2} ight)$.Substituting these into the function,we get $f(x) = 	an^{-1}\left( \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} ight) = 	an^{-1}\left( 	an\left( \frac{x}{2} ight) ight)$.Thus,$f(x) = \frac{x}{2}$.Differentiating with respect to $x$,we get $f'(x) = \frac{1}{2}$.Therefore,$f'\left( \frac{\pi}{3} ight) = \frac{1}{2}$.

If $f(x) = \tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$,then $f'\left( \frac{\pi}{3} \right) = $

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