# Solve this

Question:

If $y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right), x>0$, prove that $\frac{d y}{d x}=\frac{4}{1+x^{2}}$

Solution:

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

Using, $\sec ^{-1} x=\frac{1}{\cos ^{-1} x}$

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

Put $x=\tan \theta$

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

Using, $\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\tan 2 \theta$ and $\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$

$y=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta)$

Considering the limits

$0$0<\tan \theta<\infty0<\theta<\frac{\pi}{2}0<2 \theta<\pi$Now,$y=2 \theta+2 \thetay=4 \thetay=4 \tan ^{-1} x$Differentiating w.r.t$x$we get$\frac{d y}{d x}=\frac{d}{d x}\left(4 \tan ^{-1} x\right)\frac{d y}{d x}=\frac{4}{1+x^{2}}\$