Question

Derivative of $si{n}^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$ with respect to $co{s}^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ is

• A. $1$
• B. $cott$
• C. $tant$
• 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 $z={\cos }^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$
$={\cos }^{-1}\left(\frac{1}{\sqrt{1+{\tan }^2\theta }}\right)$
$={\cos }^{-1}(\cos \theta )$
$=\theta ={\tan }^{-1}t$
$\therefore \frac{dy}{dz}=\frac{\frac{dy}{dt}}{\frac{dz}{dt}}$
$=\frac{\frac{1}{\left(1+t^2\right)}}{\frac{1}{\left(1+t^2\right)}}=1$