Question

$sinx=\frac{2t}{1+t^2},tany=\frac{2t}{1-t^2}$

Find $\frac{dy}{dx}$

Answer 1

$sinx=\frac{2t}{1+t^2}\Rightarrow x=2ta{n}^{-1}t$

$tany=\frac{2t}{1-t^2}\Rightarrow y=2ta{n}^{-1}t$
$\therefore \frac{dx}{dt}=\frac{dy}{dt}=\frac{2}{1+t^2}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{1+t^2}}{\frac{2}{1+t^2}}=1$