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

1

We know that

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

and,

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

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

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

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

$= sin \begin{Bmatrix}\frac{\pi}{2}-2tan^{-1}x+2tan^{-1}x\end{Bmatrix}$ if $0 ≤ x ≤ 1$

$= sin \frac{\pi}{2}= 1 $