Question

If $sinx=\frac{2t}{1+t^2},tany=\frac{2t}{1-t^2},$ then $\frac{dy}{dx}$ is equal to

• A. -1
• B. 2
• C. 0
• D. 1

Answer 1

The correct option is D 1

$\sin x=\frac{2t}{1+t^2}$
Differentiate both sides w.r.t. t
$\cos x\frac{dx}{dt}=\frac{(1+t^2)\left(2\right)-2t(0+2t)}{{(1+t^2)}^2}$
$\cos x\frac{dx}{dt}=\frac{2+2t^2-4t^2}{{(1+t^2)}^2}$
$\cos x\frac{dx}{dt}=\frac{2-2t^2}{{(1+t^2)}^2}$
$\frac{dx}{dt}=\frac{2(1-t^2)}{{(1+t^2)}^2}\times \frac{1+t^2}{\sqrt{{(1+t^2)}^2-4t^2}}[\because \cos x=\sqrt{1-{\sin }^{2}x}]$
$\frac{dx}{dt}=\frac{2(1-t^2)}{{(1+t^2)}^2}\times \frac{1+t^2}{\sqrt{1+t^4+2t^2-4t^2}}$
$\frac{dx}{dt}=\frac{2(1-t^2)}{1+t^2}\times \frac{1}{\sqrt{{(1+t^2)}^2}}=\frac{2}{1+t^2}⟶\left(1\right)$
Now, $\tan y=\frac{2t}{1-t^2}$
${\sec }^{2}y\frac{dy}{dt}=\frac{(1-t^2)\left(2\right)-2t(0-2t)}{{(1-t^2)}^2}=\frac{2+2t^2}{{(1-t^2)}^2}$
$\frac{dy}{dt}=\frac{2(1+t^2)}{{(1-t^2)}^2}\times \frac{1}{[1+{\left[\frac{2t}{1-t^2}\right]}^2]}[\because {\sec }^{2}x={\tan }^{2}x+1]$
$\frac{dy}{dt}=\frac{2(1+t^2)}{{(1-t^2)}^2}\times \frac{{(1-t^2)}^2}{{(1-t^2)}^2+4t^2}=\frac{2(1-t^2)}{{(1-t^2)}^2}$
$=\frac{2}{1+t^2}⟶\left(2\right)$
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2/(1-t^2)}{2/(1-t^2)}$
$=1$