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
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),
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2}{1+t^{2}} \times \frac{1+t^{2}}{2}$
$\Rightarrow \frac{d y}{d x}=1$