Question

If ${\cos }^{-1}x>{\sin }^{-1}x,$ then $x$ lies in the interval

• A. $\left(\frac{1}{2},1\right]$
• B. ( 0, 1 ]
• C. $\left[-1\frac{1}{\sqrt{2}}\right)$
• D. [ - 1, 1 ]
• E. [ 0, 1 ]

Answer 1

Answer: (C)

We know that, ${\cos }^{-1}X$ and ${\sin }^{-1}x$ exist for $X\in [-1,1]$
Now, ${\cos }^{-1}X>{\sin }^{-1}X$
$\Rightarrow {\cos }^{-1}x>\frac{\pi }{2}-{\cos }^{-1}x$
$\Rightarrow 2{\cos }^{-1}x>\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}x>\frac{\pi }{4}$
$\Rightarrow x\in \left[-1,\frac{1}{\sqrt{2}}\right)$