Question

If $x$ takes non-positive permissible value, then ${\sin }^{-1}x=$

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

Answer 1

The correct option is B

$-{\cos }^{-1}\sqrt{1-x^2}$

Explanation for the correct option.

We know that, for $x>0$, ${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}$

We will prove it now.

Let $x=\sin \theta$

So,

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

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

Hence, option B is correct.