If |a| < 1, |b| < 1 and |x| < 1, then the solution of $ sin^{-1}\left(\frac{2a}{1+a^2}\right)-cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = tan^{-1} \left(\frac{2x}{1-x^2}\right)$, is 

$\frac{a-b}{1+ab}$

We have,

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

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

$⇒ tan^{-1} a - tan^{-1} b = tan^{-1}x $

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

$⇒ x =\frac{a-b}{1+ab}$