Question

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

Answer 1

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

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

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

${\tan }^{-1}\left(\frac{2+2x(1-x)}{-4x}\right)+{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

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

We know that,

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

$\Rightarrow -{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=0$