# Find the values of each of the following:    13. $\tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0$ and $xy<1$

Taking the value $x = \tan \Theta$  or  $\tan^{-1}x = \Theta$  and $y = \tan \Theta$  or  $\tan^{-1} y = \Theta$then we have,

$\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ]$,

$\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]$

$\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]$

$\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]$

Then,

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

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

$=\frac{x+y}{1-xy}$  Ans.

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