If $0 ≤ x ≤1$, then $tan \begin{Bmatrix}\frac{1}{2}sin^{-1}\frac{2x}{1+x^2}+\frac{1}{2}cos^{-1}\frac{1-x^2}{1+x^2}\end{Bmatrix}$ is equal to 

$\frac{2x}{1-x^2}$

We  know that

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

and,

$cos^{-1}\left(\frac{1-x^2}{1-x^2}\right)= 2 tan^{-1}x,$ if 0 ≤ x ≤ ∞

$∴ tan \begin{Bmatrix} \frac{1}{2}sin^{-1}\frac{2x}{1+x^2}+\frac{1}{2}cos^{-1}\frac{1-x^2}{1+x^2}\end{Bmatrix}$

$= tan (2 tan^{-1}x) $ for 0 ≤ x ≤ 1

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