If \( \sin x=\frac{2 t}{1+t^{2}}, \tan y=\frac{2 t}{1-t^{2}} \), then \( \frac{d y}{d x} \) is equal to

Given that,
\( \sin x=\frac{2 t}{1+t^{2}} \) and \( \tan y=\frac{2 t}{1-t^{2}} \)
So, \( x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right) \rightarrow(1) \)
and \( y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) \rightarrow(2) \)
We know that \( \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)=2 \tan ^{-1} t \) and \( \tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)=2 \tan ^{-1} t \)
So, Eqs. (1) and (2) becomes
\( x=2 \tan ^{-1} t \) and \( y=2 \tan ^{-1} t \)
Differentiating Eqs. (1) and (2) with respect to t, we get
\( \frac{d y}{d x}=\frac{d\left(\frac{2 \tan ^{-1} \theta}{d \theta}\right)}{\frac{d\left(2 \tan ^{-1} \theta\right)}{d \theta}}=1 \)