Question

Write the simplest form of ${tan}^{-1}\left[\sqrt{\frac{1-cosx}{1+cosx}}\right]$

Answer 1

We know that $\cos x=2{\cos }^2\frac{x}{2}-1=1-2{\sin }^2\frac{x}{2}$

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

$=\frac{{\sin }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}}$

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

$=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

$=\tan \frac{x}{2}$

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

$=\frac{x}{2}$