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Question:

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

Solution:

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

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

$\Rightarrow x=2 \tan ^{-1} t$ and $y=2 \tan ^{-1} t$

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

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