Question

If \[x = \sin^{- 1} \left( \frac{2 t}{1 + t^2} \right) \text{ and y } = \tan^{- 1} \left( \frac{2 t}{1 - t^2} \right), - 1 < t < 1\] porve that \[\frac{dy}{dx} = 1\] ?

 

Answer 1

\[\text{ We have, x } = \sin^{- 1} \left( \frac{2t}{1 + t^2} \right)\]

\[\text {Put t } = \tan\theta\]

\[ \Rightarrow - 1 < \tan\theta < 1\]

\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}\]

\[ \Rightarrow - \frac{\pi}{2} < 2\theta < \frac{\pi}{2}\]

\[ \therefore x = \sin^{- 1} \left( \frac{2 \tan\theta}{1 + \tan^2 \theta} \right)\]\[ \Rightarrow x = \sin^{- 1} \left( \sin2\theta \right)\]

\[ \Rightarrow x = 2\theta .......\left[ \because - \frac{\pi}{2} < 2\theta < \frac{\pi}{2} \right]\]

\[ \Rightarrow x = 2\left( \tan^{- 1} t \right) .........\left[ \because t = \sin\theta \right]\]

\[\Rightarrow \frac{dx}{dt} = \frac{2}{1 + t^2} . . . \left( i \right)\]
\[\text { Now, y } = \tan^{- 1} \left( \frac{2t}{1 - t^2} \right)\]
\[\text { put t } = \tan\theta\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{2 \tan\theta}{1 - \tan^2 \theta} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \tan 2\theta \right) \]
\[ \Rightarrow y = 2\theta .......\left[ \because - \frac{\pi}{2} < 2\theta < \frac{\pi}{2} \right]\]
\[ \Rightarrow y = 2 \tan^{- 1} t .....\left[ \because t = \tan\theta \right]\]

\[\Rightarrow \frac{dy}{dt} = \frac{2}{1 + t^2} . . . \left( ii \right)\]
\[\text { Dividing equation } \left( ii \right) \text { by } \left( i \right), \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{1 + t^2} \times \frac{1 + t^2}{2}\]
\[ \Rightarrow \frac{dy}{dx} = 1\]