Question

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

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

Answer 1

Answer: (C)

$[-1,\frac{1}{\sqrt{2}})$

We know that,

${\cos }^{-1}x$ and ${\sin }^{-1}x$ exist for $x\in [-1,1]$.

Now, ${\cos }^{-1}x>{\sin }^{-1}x$
$\Rightarrow {\cos }^{-1}x>\frac{\pi }{2}-{\cos }^{-1}x$
$\Rightarrow 2{\cos }^{-1}x>\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}x>\frac{\pi }{4}$
$\Rightarrow x\in [-1,\frac{1}{\sqrt{2}})$