Question

\[\sin x = \frac{2t}{1 + t^2}, \tan y = \frac{2t}{1 - t^2}, \text { find }  \frac{dy}{dx}\] ?

Answer 1

\[\sin x = \frac{2t}{1 + t^2}\text {  and } \tan y = \frac{2t}{1 - t^2}\]
\[ \Rightarrow x = \sin^{- 1} \frac{2t}{1 + t^2} \text { and y } = \tan^{- 1} \frac{2t}{1 - t^2}\]
\[ \Rightarrow x = 2 \tan^{- 1} t \text { and y } = 2 \tan^{- 1} t\]
\[ \Rightarrow \frac{dx}{dt} = \frac{2t}{1 + t^2} \text { and } \frac{dy}{dt} = \frac{2t}{1 + t^2}\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2t}{1 + t^2}}{\frac{2t}{1 + t^2}} = 1\]