Question

Find the value of the following: $\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]$, $|x|<1, y>0 \text { and } x y<1 \quad \text { or } \quad \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}$

Answer 1

SOLUTION : $\left.\quad \tan \frac{1}{2} \mid \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right]$,

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

L.H.S. $=\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$

$=\tan ^{-1}\left(\frac{\frac{7}{10}}{1-\frac{1}{10}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\tan ^{-1}\left(\frac{7}{9}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\tan ^{-1}\left(\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9} \cdot \frac{1}{8}}\right)=\tan ^{-1}\left(\frac{\frac{65}{72}}{\frac{65}{72}}\right)=\tan ^{-1}(1) \\$

$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}=\text { R.H.S. }$