Question

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

Answer 1

${\tan }^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ and $x<\pi$

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

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