Question

Evaluate:
$\tan \frac{1}{2}[{\sin }^{-1}\frac{2y}{1+y^2}+{\cos }^{-1}\frac{1-y^2}{1+y^2}]$

Answer 1

let $y=\tan x⟹x={\text{tan}}^{-1}y$

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

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

$\tan \frac{1}{2}[{\text{sin}}^{-1}\frac{2y}{1+y^2}+{\text{cos}}^{-1}\frac{1-y^2}{1+y^2}]=\tan \frac{1}{2}[2{\text{tan}}^{-1}y+2{\text{tan}}^{-1}y]=\tan \left(2{\text{tan}}^{-1}y\right)=\frac{2y}{1-y^2}$