Question

$ta{n}^{-1}\sqrt{x}=\frac{1}{2}co{s}^{-1}\left(\frac{1-x}{1+x}\right),xϵ[0,1].$

Answer 1

Given, $ta{n}^{-1}\sqrt{x}=\frac{1}{2}co{s}^{-1}\left(\frac{1-x}{1+x}\right),xϵ[0,1].$
$LHS=ta{n}^{-1}\sqrt{x}=\frac{1}{2}\times \left(2ta{n}^{-1}\sqrt{x}\right)=\frac{1}{2}\times co{s}^{-1}\frac{1-(\sqrt{x}{)}^2}{1+(\sqrt{x}{)}^2}[\because 2ta{n}^{-1}x=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)]=\frac{1}{2}co{s}^{-1}\left(\frac{1-x}{1+x}\right)=RHS$