Inverse Trigonometric Functions

If \(\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) where \(a, x \in(0,1)\), then the value of \(x\) is