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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),

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

$\Rightarrow \frac{d y}{d x}=1$

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