The value of $tan\frac{1}{2}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}\end{Bmatrix}$ if $|x|<1, y > 0 $ and $xy < 1$, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{x+y}{1-xy}$
$\tan\frac{1}{2}\left\{\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right\}$
as $\sin^{-1}\frac{2x}{1+x^2}=θ$
$\tan^{-1}\frac{2x}{1+x^2}=θ$
$\cos^{-1}\frac{1-y^2}{1+y^2}=\phi$
$\tan^{-1}\frac{2y}{1+y^2}=\phi⇒\tan\frac{1}{2}\left(\tan^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right)$
$=\tan\frac{1}{2}(2\tan^{-1}x+2\tan^{-1}y)$
$=\frac{x+y}{1-xy}$