Question

Find the value of $\tan \frac{1}{2}[{\sin }^{-1}\frac{2x}{1+x^2}+{\cos }^{-1}\frac{1-y^2}{1+y^2}],\left|x\right|<1,y>0$ and $xy<1$

Answer 1

Let $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$

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

Let $y=\tan \varphi \Rightarrow \varphi ={\tan }^{-1}y$

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

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

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

$=\tan [{\tan }^{-1}x+{\tan }^{-1}y]$

$=\tan \left[{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right]=\frac{x+y}{1-xy}$