# Solve this

Question:

If $x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$ and $y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right),-1 Solution: We have,$x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$Put$t=\tan \theta\Rightarrow-1<\tan \theta<1\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}\Rightarrow-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}\therefore x=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\Rightarrow x=\sin ^{-1}(\sin 2 \theta)\Rightarrow x=2 \theta\left[\because-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}\right]\Rightarrow x=2\left(\tan ^{-1} t\right)[\because t=\sin \theta]\Rightarrow \frac{d x}{d t}=\frac{2}{1+t^{2}}$......(1) Now,$y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)$put$t=\tan \theta\Rightarrow y=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)\Rightarrow y=\tan ^{-1}(\tan 2 \theta)\Rightarrow y=2 \theta\left[\because-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}\right]\Rightarrow y=2 \tan ^{-1} t[\because t=\tan \theta]\Rightarrow \frac{d y}{d t}=\frac{2}{1+t^{2}}$......(2) Dividing equation (ii) by (i),$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2}{1+t^{2}} \times \frac{1+t^{2}}{2}\Rightarrow \frac{d y}{d x}=1\$