Question

If ${\cos }^{-1}x>{\sin }^{-1}x$, then

• A. $\frac{1}{\sqrt{2}}<x<1$
• B. $0\le x<\frac{1}{\sqrt{2}}$
• C. $-1\le x<\frac{1}{\sqrt{2}}$
• D. $x>0$

Answer 1

Answer: (C)

Given that, ${\cos }^{-1}x>{\sin }^{-1}x$

Then, ${\sin }^{-1}x={\cos }^{-1}x\Rightarrow x=\frac{1}{\sqrt{2}}$
From the figure, it is clear that $-1\le x<\frac{1}{\sqrt{2}}$.