Question

If $x$ takes negative permissible value, then ${\sin }^{-1}x$ is equal to

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

Answer 1

The correct option is B $-{\cos }^{-1}\sqrt{1-x^2}$

$si{n}^{-1}\left(x\right)$
$=co{s}^{-1}\left(\sqrt{1-x^2}\right)$ , for $x>0$
Since x takes negative permissible value.
$si{n}^{-1}\left(x\right)=-co{s}^{-1}\left(\sqrt{1-x^2}\right)$