Question

If $\tan x=\frac{2t}{1-t^2}$ and $\sin y=\frac{2t}{1+t^2}$, then the value of $\frac{dy}{dx}$ is

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

Answer 1

The correct option is A $1$

Given,

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

Now,

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

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

From Eq. (i),
$\frac{dx}{dt}=\frac{2}{1+t^2}$

From Eq. (ii),
$\frac{dy}{dt}=\frac{2}{1+t^2}$

$\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2}{(1+t^2)}\times \frac{(1+t^2)}{2}=1$