# Write the following functions in the simplest form:    7. $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

Given that $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

We have in inside the root the term : $\frac{1-\cos x}{1 + \cos x}$

Put $1-\cos x = 2\sin^2\frac{x}{2}$   and    $1+\cos x = 2\cos^2\frac{x}{2}$,

Then we have,

$=\tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} \right )$

$=\tan^{-1}\left( \frac{\sin \frac{x}{2}}{\cos\frac{x}{2}} \right )$

$=\tan^{-1}(\tan\frac{x}{2}) = \frac{x}{2}$

Hence the simplest form is $\frac{x}{2}$

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