Question

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

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