Question

Find the value of $\tan \left\{\frac{1}{2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+\frac{1}{2}{\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$; if $x>y>1$.

Answer 1

We have,

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

Put $x=y=\tan \theta$

Then, $\theta ={\tan }^{-1}x={\tan }^{-1}y$

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

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

$=\tan \{\frac{1}{2}{\sin }^{-1}\sin 2\theta +\frac{1}{2}{\cos }^{-1}\cos 2\theta \}$

$=\tan \{\frac{1}{2}\times 2\theta +\frac{1}{2}\times 2\theta \}$

$=\tan 2\theta$

$=\frac{2\tan \theta }{1-{\tan }^2\theta }$

$=\frac{2x}{1-x^2}=\frac{2y}{1-y^2}$

Hence, this is the answer.