${\sin }^{-1}\frac{2a}{1+a^2}+{\cos }^{-1}\frac{1-b^2}{1+b^2}=2{\tan }^{-1}x$.

We have,

${\sin }^{-1}\frac{2a}{1+a^2}+{\cos }^{-1}\frac{1-b^2}{1+b^2}=2{\tan }^{-1}x$

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

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

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

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

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

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

Hence, this is the answer.