Question

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

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

Answer 1

The correct option is D $-1\le x<\frac{1}{\sqrt{2}}$

For ${\sin }^{-1}$ and ${\cos }^{-1}x$ to be real -1 $\le$ x < 1
But ${\cos }^{-1}x>{\sin }^{-1}x$
$\Rightarrow 2{\cos }^{-1}x>\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}x>\frac{\pi }{4}$
Or $x<\frac{1}{\sqrt{2}}$
$\Rightarrow -1\le x<\frac{1}{\sqrt{2}}$