Question

Prove: ${\tan }^{-1}\sqrt{x}=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right),x\in [0,1]$

Answer 1

Let $x={\tan }^2\theta$

$\Rightarrow \sqrt{x}=\tan \theta$

$\Rightarrow \theta ={\tan }^{-1}\sqrt{x}$

$\therefore \frac{1-x}{1+x}=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$

Now, we have: $R.H.S.=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)$

$=\frac{1}{2}{\cos }^{-1}\left(\cos 2\theta \right)$

$=\frac{1}{2}\times 2\theta =\theta$

$={\tan }^{-1}\sqrt{x}=L.H.S.$