Question

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

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

Answer 1

Answer: (E)

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
$y={\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)$
$={\tan }^{-1}(\tan 2\theta )=2\theta$
$=2{\tan }^{-1}t....(ii)$
From Eqs. (i) and (ii), $y=x$
$\Rightarrow \frac{dy}{dx}=1$