#### Question

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

#### Solution

\[\sin x = \frac{2t}{1 + t^2}\text { and } \tan y = \frac{2t}{1 - t^2}\]

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

\[ \Rightarrow x = 2 \tan^{- 1} t \text { and y } = 2 \tan^{- 1} t\]

\[ \Rightarrow \frac{dx}{dt} = \frac{2t}{1 + t^2} \text { and } \frac{dy}{dt} = \frac{2t}{1 + t^2}\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2t}{1 + t^2}}{\frac{2t}{1 + t^2}} = 1\]

Is there an error in this question or solution?

Solution for question: Sin X = 2 T 1 + T 2 , Tan Y = 2 T 1 − T 2 , Find D Y D X ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)