Question

If $u={\tan }^{-1}\left(\frac{\sqrt{1-x^2}-1}{x}\right)$ and $v={\sin }^{-1}x$, then $\frac{du}{dv}$ is equal to

• A. $\sqrt{1-x^2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-x$
• E. $-2$

Answer 1

Answer: (B)

$-\frac{1}{2}$

Given, $u={\tan }^{-1}\left\{\frac{\sqrt{1-x^2}-1}{x}\right\}$
and $v={\sin }^{-1}x$
Put $x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$, then
$u={\tan }^{-1}\left\{\frac{\cos \theta -1}{\sin \theta }\right\}(\because 1-{\cos }^2\theta ={\sin }^2\theta )$
$=-{\tan }^{-1}\left\{\frac{1-\cos \theta }{\sin \theta }\right\}$
$=-{\tan }^{-1}\left\{\frac{2{\sin }^2\theta /2}{2\sin \theta /2\cdot \cos \theta /2}\right\}$
$=-{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$
$=-\frac{1}{2}\theta =-\frac{1}{2}{\sin }^{-1}x$
Therefore, $\frac{du}{dx}=-\frac{1}{2\sqrt{1-x^2}}$
and $\frac{dv}{dx}=\frac{1}{\sqrt{1-x^2}}$
Thus $\frac{du}{dv}=\frac{du}{dx}\times \frac{dx}{dv}=-\frac{1}{2\sqrt{1-x^2}}\times \sqrt{1-x^2}$
$=-\frac{1}{2}$