Question

$\tan \frac{1}{2}[{\sin }^{-1}\frac{2x}{1+x^2}+{\cos }^{-1}\frac{1-y^2}{1+y^2}]$, $|x|<1,y>0$ and $xy<1$.

Answer 1

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

$=\tan \frac{1}{2}[{\sin }^{-1}\frac{2\tan \alpha }{1+{\tan }^2\alpha }+{\cos }^{-1}\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }]$

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

$=\tan \frac{1}{2}[2\alpha +2\beta ]$ Let $x=\tan \alpha ;y=\tan \beta$

$=\tan (\alpha +\beta )$

$⟹\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }=\frac{x+y}{1-xy}$