Question

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

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

Answer 1

Answer: (C)

$cos^{-1}x > sin^{-1}x$
$\Rightarrow \frac{\pi}{2} - sin^{-1}x > sin^{-1}x$
$\Rightarrow \frac{\pi }{2} > 2\, sin^{-1}x $
$\Rightarrow sin^{-1}x < \frac{\pi }{4}$
$\Rightarrow x < sin \frac{\pi }{4}$
$\Rightarrow x < \frac{1}{\sqrt{2}}$