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

Answer: (A)

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