Question

Prove that: ${\tan }^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right]=\frac{x}{2}$

Answer 1

Take = $cosx=\frac{1-ta{n}^{2}x/2}{1+ta{n}^{2}x/2}$

$\frac{1-cosx}{1+cosx}=\frac{1-\frac{1-ta{n}^{2}x/2}{1+ta{n}^{2}x/2}}{1+\frac{1-ta{n}^{2}x/2}{1+ta{n}^{2}x/2}}$

$=\frac{1+ta{n}^{2}x/2-1+ta{n}^{2}x/2}{1+ta{n}^{2}x/2+1-ta{n}^{2}x/2}$

$=\frac{2ta{n}^{2}x/2}{2}=ta{n}^{2}x/2$

$\sqrt{\frac{1-cosx}{1+cosx}}=\sqrt{ta{n}^{2}x/2}=tanx/2$

Now, $ta{n}^{-1}\left[tan\right(\frac{x}{2}\left)\right]=\frac{x}{2}$