question_answer

Prove that \[\left\{ \frac{1}{2}{{\sin }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)+\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right\}\] \[=\left( \frac{x+y}{1-xy} \right),\] where |x|<1, y > 0 and xy < 1.

Answer:

Putting\[x=\tan \theta \]and \[y=\tan \phi ,\]we get \[LHS=\tan \left\{ \frac{1}{2}{{\sin }^{-\,1}}\left( \frac{2x}{1-{{x}^{2}}} \right)+\frac{1}{2}{{\cos }^{-\,1}}\left( \frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right\}\] \[=\tan \left\{ \frac{1}{2}{{\sin }^{-\,1}}\left( \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)+\frac{1}{2}{{\cos }^{-\,1}}\left( \frac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right) \right\}\] \[=\tan \left\{ \frac{1}{2}{{\sin }^{-\,1}}(\sin 2\theta )+\frac{1}{2}{{\cos }^{-\,1}}(\cos 2\phi ) \right\}\] \[=\tan \left\{ \left( \frac{1}{2}\times 2\theta  \right)+\left( \frac{1}{2}\times 2\phi  \right) \right\}\] \[=\tan \,(\theta +\phi )=\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }=\frac{(x+y)}{(1-xy)}=RHS\] \[\therefore \] \[\tan \left\{ \frac{1}{2}{{\sin }^{-\,1}}\left( \frac{2x}{1-{{x}^{2}}} \right)+\frac{1}{2}{{\cos }^{-\,1}}\left( \frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right\}=\frac{x+y}{1-xy}\]