Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $\infty$

Answer 1

Answer: (B)

Given, $y={\sin }^{-1}\left(\frac{2t}{1+t^2}\right),x={\tan }^{-1}\left(\frac{2t}{1-t^2}\right)$
Let $t=\tan (\varphi /2)$
$\therefore \frac{2t}{1+t^2}=\frac{2\tan (\varphi /2)}{1+{\tan }^2(\varphi /2)}=2(\varphi /2)=\sin \varphi$
$\Rightarrow \frac{2t}{1+t^2}=\sin \varphi ?$ (i) Also, $\frac{2t}{1-t^2}=\frac{2\tan (\varphi /2)}{1-{\tan }^2(\varphi /2)}$ $=\tan 2(\varphi /2)=\tan \varphi$ ..(ii)
From (i) and (ii),
$y={\sin }^{-1}\sin \varphi$ and $x={\tan }^{-1}\tan (\varphi )$
$\Rightarrow$$y=\varphi ,x=\varphi \Rightarrow \frac{dy}{d\varphi }=1,\frac{dx}{d\varphi }=1\therefore \frac{dy}{dx}=1$
Alternate Method: Given,
$y={\sin }^{-1}\left(\frac{2t}{1+t^2}\right),x={\tan }^{-1}\left(\frac{2t}{1-t^2}\right)$
$\Rightarrow$
$\frac{dy}{dt}=\frac{1}{\sqrt{1-{\left(\frac{2t}{1+t^2}\right)}^2}}\times \frac{d}{dt}\left(\frac{2t}{1+t^2}\right)$
$=\frac{1+t^2}{1-t^2}\left[\frac{(1+t{)}^2\times 2-2t(2t)}{{\left(1+t^2\right)}^2}\right]=\frac{2}{1+t^2}$ .. (i) and
$\frac{dx}{dt}=\frac{1}{1+{\left(\frac{2t}{1-t^2}\right)}^2}\frac{d}{dt}\left(\frac{2t}{1-t^2}\right)=\frac{1}{1+\left(\frac{2t}{1-t^2}\right)}\frac{\left(1-t^2\right)\times 2+2t(2t)}{{\left(1-t^2\right)}^2}$
$=\frac{{\left(1-t^2\right)}^2}{{\left(1+t^2\right)}^2}\left[\frac{\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\right]=\frac{2}{1+t^2}$ .. (ii)
$\therefore$
$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2}{1+t^2}\times \frac{1+t^2}{2}=1$ [from (i) and (ii)]