Q72 / 120 · VITEEE 2009 · Mathematics
$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$,
$y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right) \Rightarrow \frac{d y}{d x}$ is equal to
- A0
- B$\tan t$
- Correct answerC1
- D$\sin t \cos t$
Solution
We have, $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$
and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$
$$
\Rightarrow x=\tan ^{-1} t \text { and } y=\tan ^{-1} t
$$
$$
\therefore y=x \Rightarrow \frac{d y}{d x}=1
$$
Hence, the correct answer is option: 'C'.