Question

sinx=2t/1+t^2,tany=2t/1-t^2 find dy/dx​

Answer 1

HERE'S YOUR ANSWER IN THE ATTACHMENT

Answer 2

dy/dx = 1

• Given,

             $sin x = \frac{2t}{1+t^{2} }$  --------(1)   $tan y = \frac{2t}{1-t^{2} }$   ---------------(2)

• Let

           $t = tan \alpha$ ⇒ $\alpha = tan^{-1} t$

• Substituting t in (1) and (2), we get

            $sin x = \frac{2tan\alpha }{1+tan^{2}\alpha }$               $tan y = \frac{2tan\alpha }{1 - tan^{2}\alpha }$

       ⇒ $sin x = sin 2\alpha$              ⇒ $tan y = tan 2\alpha$

       ⇒ $x = 2\alpha$                        ⇒ $y = 2\alpha$

• Differentiating on both sides w.r.t $\alpha$

         ⇒ $\frac{dx}{d\alpha } = 2$   -----------(3)   ⇒ $\frac{dy}{d\alpha } = 2$ ------------------(4)

• (4)/(3) ⇒ $(\frac{dy}{d\alpha } )/(\frac{dx}{d\alpha } ) = \frac{2}{2}$

                 ⇒  dy/dx = 1

NOTE: I used alpha in making solution as there is no option for theta.

But you must use theta not alpha.