Question

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

Answer 1

$R.H.S.=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)$
Put $x={\tan }^2\theta$ in given equation
$=\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
$=\frac{1}{2}{\cos }^{-1}\left(\frac{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}\right)$
$=\frac{1}{2}{\cos }^{-1}\left(\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta +{\sin }^2\theta }\right)$
$=\frac{1}{2}{\cos }^{-1}\left(\frac{{\cos }^2\theta -{\sin }^2\theta }{1}\right)$
$=\frac{1}{2}{\cos }^{-1}\left(\cos 2\theta \right)$
$=\frac{1}{2}\times 2\theta =\theta \cdots \left(i\right)$

$\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)=\theta$
$x={\tan }^2\theta$
$\Rightarrow \sqrt{x}=\tan \theta$
$\Rightarrow {\tan }^{-1}\sqrt{x}=\theta \cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$
${\tan }^{-1}\sqrt{x}=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)$
Hence proved.