If  ${\cos }^{-1}\left(\frac{1-a^2}{1+a^2}\right)-{\cos }^{-1}\left(\frac{1-b^2}{1+b^2}\right)=2{\tan }^{-1}x$, then ‘x’ is equal to

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

$\therefore 2{\tan }^{-1}a-2{\tan }^{-1}b=2{\tan }^{-1}x$

$\Rightarrow {\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}x$

$\Rightarrow {\tan }^{-1}\left(\frac{a-b}{1+ab}\right)={\tan }^{-1}x$

$\therefore x=\frac{a-b}{1+ab}$