Question

When $ y={{\sin }^{-1}}\left( \frac{2t}{1+{{t}^{2}}} \right) $ , $ x={{\tan }^{-1}}\left( \frac{2t}{1-{{t}^{2}}} \right) $ where $ -1

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

Answer 1

Answer: (B)

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