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, $\sin x=\frac{2 t}{1+t^{2}}$
$\Rightarrow x=\sin ^{-1} \frac{2 t}{1+t^{2}}$
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{2 t}{1-t^{2}} $
$\Rightarrow y =\tan ^{-1} \frac{2 t}{1-t^{2}}$
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), $y=x$
$\Rightarrow \frac{d y}{d x}=1$