Question

Find the value of the expression $\tan \frac{1}{2}\left[{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right]$.Where $|x|<1$ and $y>0$ such that $xy<1$.

Answer 1

Applying the identities of trigonometry in $\tan \frac{1}{2}\left[{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right]$.

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

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

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

$=\frac{x+y}{1-xy}$