Question

If $\left|t\right|<1$, $\sin x=\frac{2t}{1+t^2}$, $\tan y=\frac{2t}{1-t^2}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{x}$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $-\frac{1}{x}$
• E. $1$

Answer 1

Answer: (E)

$1$

We have, $\sin x=\frac{2t}{1+t^2}$
$\Rightarrow x={\sin }^{-1}\frac{2t}{1+t^2}$
On putting $t=\tan \theta$, we get
$x={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$
$={\sin }^{-1}\sin 2\theta =2\theta$
$\Rightarrow x=2{\tan }^{-1}t$ ....(i)
Also, $\tan y=\frac{2t}{1-t^2}$
$\Rightarrow y={\tan }^{-1}\frac{2t}{1-t^2}$
On putting $t=\tan \theta$, we get
$y={\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)$
$={\tan }^{-1}\left(\tan 2\theta \right)=2\theta$
$=2{\tan }^{-1}t$ ....(ii)
From equations (i) and (ii), $y=x$
$\Rightarrow \frac{dy}{dx}=1$