If  tany=(2t)/(1-t^2), sinx=(2t)/(1+t^2) then  (dy)/(dx)=?    

দেওয়া আছে,

\(\tan y = \frac{2t}{1-t^2}\)

\(\sin x = \frac{2t}{1+t^2}\)

আমরা জানি,

\(\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}\)

এবং

\(\sin 2\theta = \frac{2\tan \theta}{1+\tan^2 \theta}\)

ধরি, \(t = \tan \theta\)

তাহলে,

\(\tan y = \frac{2\tan \theta}{1-\tan^2 \theta} = \tan 2\theta\)

\(\implies y = 2\theta\)

এবং

\(\sin x = \frac{2\tan \theta}{1+\tan^2 \theta} = \sin 2\theta\)

\(\implies x = 2\theta\)

সুতরাং, \(y = x\)

অতএব,

\(\frac{dy}{dx} = \frac{d}{dx}(x) = 1\) 🎉