Q84 / 150 · MHT CET 2012 · Mathematics
If $\tan x=\frac{2 t}{1-t^{2}}$ and $\sin y=\frac{2 t}{1+t^{2}}$, then the value of $\frac{d y}{d x}$ is
- Correct answerA1
- B$t$
- C$\frac{1}{1-t}$
- D$\frac{1}{1+t}$
Solution
Given, $\tan x=\frac{2 t}{1-t^{2}}$ and $\sin y=\frac{2 t}{1+t^{2}}$
Now, $x=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)$
$x=2 \tan ^{-1} t$..(i)
and $\quad y=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$
$y=2 \tan ^{-1} t$...(ii)
From Eq. (i), $\frac{d x}{d t}=\frac{2}{1+t^{2}}$
From Eq. (ii), $\frac{d y}{d t}=\frac{2}{1+t^{2}}$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{2}{\left(1+t^{2}\right)} \times \frac{\left(1+t^{2}\right)}{2}=1$
Hence, the correct answer is option: 'A'.