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

Answer

$\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan^{-1}\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}=\tan^{-1}\tan\frac{x}{2}=\frac{x}{2}$

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