Question

Derivative of $sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$ with respect to $cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ is

• A. $1$
• B. $cot\,t$
• C. $tan\,t$
• D. $0$

Answer 1

Answer: (A)

Let $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$
Put $t=\tan \theta \Rightarrow \theta=\tan ^{-1} t$
$=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\right)=\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)$
$=\sin ^{-1}(\sin \theta)=\theta=\tan ^{-1} t$
and $z=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$
$=\cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^{2} \theta}}\right)$
$=\cos ^{-1}(\cos \theta)$
$=\theta=\tan ^{-1} t$
$\therefore \frac{d y}{d z}=\frac{\frac{d y}{d t}}{\frac{d z}{d t}}$
$=\frac{\frac{1}{\left(1+t^{2}\right)}}{\frac{1}{\left(1+t^{2}\right)}}=1$