Question

If $tanx=\frac{2t}{1-t^2}$ and $siny=\frac{2t}{1+t^2}$ , then the value of $\frac{dy}{dx}$ is

• A. $1$
• B. $t$
• C. $\frac{1}{1-t}$
• D. $\frac{1}{1+t}$

Answer 1

Answer: (A)

Given, $\tan x=\frac{2t}{1-t^2}$ and $\sin y=\frac{2t}{1+t^2}$
Now, $x={\tan }^{-1}\left(\frac{2t}{1-t^2}\right)$
$x=2{\tan }^{-1}t\dots (i)$
and $y={\sin }^{-1}\left(\frac{2t}{1+t^2}\right)$
$y=2{\tan }^{-1}t\dots (ii)$
From Eq. (i),
$\frac{dx}{dt}=\frac{2}{1+t^2}$
From Eq. (ii),
$\frac{dy}{dt}=\frac{2}{1+t^2}$
$\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2}{\left(1+t^2\right)}\times \frac{\left(1+t^2\right)}{2}=1$