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# Find the value of

Question:

Find the value of $\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right],|x|<1, y>0$ and $x y<1$

Solution:

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

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

Let $y=\tan \Phi$. Then, $\Phi=\tan ^{-1} y$.

$\therefore \cos ^{-1} \frac{1-y^{2}}{1+y^{2}}=\cos ^{-1}\left(\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}\right)=\cos ^{-1}(\cos 2 \phi)=2 \phi=2 \tan ^{-1} y$

$\therefore \tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]$

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

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

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

$=\frac{x+y}{1-x y}$