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Find $\frac{d y}{d x}$, when

If $\sin x=\frac{2 t}{1+t^{2}}, \tan y=\frac{2 t}{1-t^{2}}$, find $\frac{d y}{d x}$


$\sin x=\frac{2 t}{1+t^{2}}, \tan y=\frac{2 t}{1-t^{2}}$

$x=\sin ^{-1} \frac{2 t}{1+t^{2}}$ and $y=\tan ^{-1} \frac{2 t}{1-t^{2}}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{\sqrt{1-\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}}\right)^{2}}} \times \frac{2\left(1+\mathrm{t}^{2}\right)-(2 \mathrm{t})(2 \mathrm{t})}{\left(1+\mathrm{t}^{2}\right)^{2}}$

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

$\frac{d y}{d t}=\frac{1}{1+\left(\frac{2 t}{1+t^{2}}\right)^{2}} \times \frac{2\left(1-t^{2}\right)-(2 t)(-2 t)}{\left(1-t^{2}\right)^{2}}$

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

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

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