If  $f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right),x>1$ then  $f\left(5\right)$ is equal to

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

$\Rightarrow x=\tan \theta \Rightarrow \tan \theta >1\Rightarrow \theta ={\tan }^{-1}\left(x\right)$

$\Rightarrow \theta \in ({45}^0,{90}^{\circ })\Rightarrow 2\theta \in ({90}^0,{180}^{\circ })$

$\Rightarrow f\left(x\right)=2\theta +{\sin }^{-1}[\sin 2\theta ]$

$=2\theta +\sin [\sin (\pi -2\theta \left)\right]=\pi$