If $x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2t}{1+t^2}\text{, then }\frac{dy}{dx}$ is equal to.

Let $x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2t}{1+t^2}$

Put, $\mathrm{t}=\tan \theta$ we get

$\begin{array}{l}x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\text{ and }y=\frac{2\tan \theta }{1+{\tan }^2\theta }\\ \Rightarrow x=\cos 2\theta \text{ and }y=\sin 2\theta \\ \therefore \frac{dx}{d\theta }=-2\sin 2\theta \text{ and }\frac{dy}{d\theta }=2\cos 2\theta \end{array}$

Now, $\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=-\frac{\cos 2\theta }{\sin 2\theta }=-\frac{x}{y}$