# Solve this

Question:

Find $\frac{d y}{d x}$, when

If $x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$ and $y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right),-1 Solution: as$\mathrm{x}=\sin ^{-1}\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}}\right)$Put$t=\tan \thetax=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)=\sin ^{-1} \sin 2 \theta=2 \theta\left[\right.$since,$\left.\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]x=2\left(\tan ^{-1} t\right)[$since,$t=\sin \theta]$differentiating it with respect to$t$,$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2}{1+\mathrm{t}^{2}}$....(1) Now,$y=\tan ^{-1} \frac{2}{1+t^{2}}$Put$t=\tan \thetay=\tan ^{-1} \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\tan ^{-1} \tan 2 \theta\left[\right.$since$\left.\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right]=2 \thetay=2 \tan ^{-1} t$[since$\left.t=\tan \theta\right]$differentiating it with respect to$t$,$\frac{d y}{d t}=\frac{2}{1+t^{2}} \cdots \cdots(2)$Dividing equation (2) by (1),$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2}{1+t^{2}} \times \frac{1+t^{2}}{2}\frac{d y}{d x}=1\$