Write the function in the simplest form: tan | vedclass

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(B) Given expression: $	an ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}ight)$Using trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and

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$\frac{x}{2}$

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(B) Given expression: $	an ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}ight)$Using trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:$= 	an ^{-1}\left(\sqrt{\frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}}ight)$$= 	an ^{-1}\left(\sqrt{	an^2 \frac{x}{2}}ight)$$= 	an ^{-1}\left(\left| 	an \frac{x}{2} ight|ight)$Since $x  0$,$	an \frac{x}{2}$ is positive,so $\left| 	an \frac{x}{2} ight| = 	an \frac{x}{2}$.$= 	an ^{-1}\left(	an \frac{x}{2}ight) = \frac{x}{2}$

Write the function in the simplest form: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), x < \pi$

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