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

$\begin{aligned} \frac{d y}{d x} & =\frac{\frac{dy}{d t}}{\frac{d x}{d t}} \\ & =\frac{\frac{d}{d t}\left(\tan ^{-1} \frac{2 t}{1-t^{2}}\right)}{\frac{d}{d t}\left(\sin ^{-1} \frac{2 t}{1+t^{2}}\right)} \\ & =\frac{2 \cdot \frac{1}{1+t^{2}}}{2 \cdot \frac{1}{1+t^{2}}}=1\end{aligned}$