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
as $\mathrm{x}=\sin ^{-1}\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}}\right)$
Put $t=\tan \theta$
$x=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$=\sin ^{-1} \sin 2 \theta$
$=2 \theta\left[\right.$ 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 $t=\tan \theta$
$y=\tan ^{-1} \frac{2 \tan \theta}{1-\tan ^{2} \theta}$
$=\tan ^{-1} \tan 2 \theta\left[\right.$ since $\left.\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right]$
$=2 \theta$
$y=2 \tan ^{-1} t$ [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),
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2}{1+t^{2}} \times \frac{1+t^{2}}{2}$
$\frac{d y}{d x}=1$