# Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $\sin x=\large\frac{2t}{1+t^2},$$\tan y=\large\frac{2t}{1-t^2} ## 1 Answer Toolbox: • To find \large\frac{dy}{dx} in the case of parametric functions,if x=\phi(t) and y=\psi(t),then \large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
Given : $\sin x=\large\frac{2t}{1+t^2}$
$\tan y=\large\frac{2t}{1-t^2}$
Consider $\sin x=\large\frac{2t}{1+t^2}$
$\Rightarrow x=\sin^{-1}\big(\large\frac{2t}{1+t^2}\big)$
Put $t=\tan\theta$
$\theta=\tan^{-1}(t)$
$\Rightarrow x=\sin^{-1}\big(\large\frac{2\tan\theta}{1+\tan^2\theta}\big)$
But $\large\frac{2\tan \theta}{1+\tan^2\theta}=$$\sin 2\theta x=\sin^{-1}(\sin 2\theta) Therefore x=2\theta Substituting for \theta we get, x=2\tan^{-1}(t) Step 2: Differentiating w.r.t to t we get, \large\frac{dx}{dt}$$=2.\large\frac{1}{1+t^2}=\frac{2}{1+t^2}$
Consider $\tan y=\large\frac{2t}{1+t^2}$
$\Rightarrow y=\tan^{-1}\big(\large\frac{2t}{1-t^2}\big)$
Put $t=\tan \theta$
$\Rightarrow \theta=\tan^{-1}(t)$
$y=\tan^{-1}\big(\large\frac{2\tan\theta}{1-\tan^2\theta}\big)$
But $\large\frac{2\tan\theta}{1-\tan^2\theta}=$$\tan 2\theta Therefore y=\tan^{-1}(\tan 2\theta) \qquad\qquad\;\;=2\theta Step 3: Substituting for \theta we get, y=2\tan^{-1}t Differentiating w.r.t t we get, \large\frac{dy}{dt}=$$2.\large\frac{1}{1+t^2}$
$\Rightarrow \large\frac{2}{1+t^2}$
$\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}=\large\frac{2/1+t^2}{2/1+t^2} Therefore \large\frac{dy}{dx}$$=1$