Question

If ${}^{\prime }{x}^{\prime }$ 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

Answer: (B)

Let ${\sin }^{-1}x=y.$ Then, $x=\sin y$
Since, $-1\le x<0$, therefore $-\frac{\pi }{2}\le {\sin }^{-1}x<0$
$\Rightarrow -\frac{\pi }{2}\le y<0$
Now, $\cos y=\sqrt{1-{\sin }^{2}y}$
$\Rightarrow \cos y=\sqrt{1-x^2}$ for $0\le y\le \pi$
But $-\frac{\pi }{2}\le y<0$
$\Rightarrow \frac{\pi }{2}\ge -y>0$
$\Rightarrow \cos (-y)=\sqrt{1-x^2}$
$\Rightarrow -y={\cos }^{-1}\sqrt{1-x^2}$
$\Rightarrow y=-{\cos }^{-1}\sqrt{1-x^2}$