Question

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

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

Answer 1

The correct option is A

$–{\cos }^{-1}\sqrt{1-x^2}$

Explanation for the correct answer:

For $x>0$, ${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}$

Let us prove this.

Let $x=\sin \theta$, then

$\begin{array}{rcl}{\cos }^{-1}\sqrt{1-x^2}& =& {\cos }^{-1}\sqrt{1-{\sin }^2\theta }\\ & =& {\cos }^{-1}\sqrt{{\cos }^2\theta }\\ & =& {\cos }^{-1}\left(\cos \theta \right)\\ & =& \theta \\ & =& {\sin }^{-1}x\end{array}$

So, for $x<0$, ${\sin }^{-1}x=-{\cos }^{-1}\sqrt{1-x^2}$

Hence, option (A) is correct.