If $\tan y=\frac{2t}{1-t^2}\text{ and }\sin x=\frac{2t}{1+t^2}\text{, then }\frac{dy}{dx}=\text{ ? }$

  $\tan y=\frac{2t}{1-t^2} ...\left(i\right)$

and $\sin x=\frac{2t}{1+t^2}$           …(ii)

From (i), differentiating w.r.t. $t$ of $y,$ we get, 

 ${\sec }^{2}y\frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}$ and $\frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\cdot \frac{1}{\left(1+{\tan }^{2}y\right)}$

or $\frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\cdot \frac{1}{\left[1+{\left.\left(\frac{2t}{1-t^2}\right)\right|}_{\rfloor }^2\right.⌉}=\frac{2}{1+t^2} ...\left(iii\right)$

and from (ii), differentiating w.r.t. t of $x,$ we get

  $\cos x\frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}$

or $\frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\frac{1}{\sqrt{1-\frac{(2t{)}^2}{{\left(1+t^2\right)}^2}}}=\frac{2}{1+t^2} ...\left(iv\right)$

Hence, $\frac{dy}{dx}=1.$