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 $t=\tan \theta$

$x=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$

$=\sin ^{-1} \sin 2 \theta$

$=2 \theta\left[\right.$ 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),

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

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

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