Question

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

Answer 1

$$\begin{gathered} \sin x=\frac{2t}{1+t^2}\mathrm{and}\tan y=\frac{2t}{1-t^2} \\ \Rightarrow x={\sin }^{-1}\frac{2t}{1+t^2}\mathrm{and}y={\tan }^{-1}\frac{2t}{1-t^2} \\ \Rightarrow x=2{\tan }^{-1}t\mathrm{and}y=2{\tan }^{-1}t \\ \Rightarrow \frac{dx}{dt}=\frac{2t}{1+t^2}\mathrm{and}\frac{dy}{dt}=\frac{2t}{1+t^2} \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{{\displaystyle \frac{2t}{1+t^2}}}{{\displaystyle \frac{2t}{1+t^2}}}=1 \end{gathered}$$