Question

If $(\cos^{-1}x)^2-(\sin^{-1}x)^2>0$, then

• A. $x<\frac{1}{2}$
• B. $-1<\,x\,\sqrt2$
• C. $0\leq\,x\,<\frac{1}{\sqrt2}$
• D. $-1\leq\,x\,<\frac{1}{\sqrt2}$

Answer 1

Answer: (D)

$\left(cos^{-1} x\right)^{2}-\left(sin^{-1}x\right)^{2}>0$
$ \Rightarrow \left(cos^{-1}x+sin^{-1}x\right)\left(cos^{-1}x -sin^{-1}x\right) > 0$
$\Rightarrow \frac{\pi}{2}\left(\frac{\pi}{2} -2\,sin^{-1}x\right)>0 $
$\Rightarrow \frac{\pi}{2}> 2 \,sin^{-1}x $
$\Rightarrow \frac{\pi}{4}>sin^{-1} x$ i.e., $sin^{-1}x < \frac{\pi}{4}$
Also $-\frac{\pi}{2}\le sin^{-1} x \le\frac{\pi}{2} $
$\therefore -\frac{\pi}{2} \le sin^{-1}x < \frac{\pi}{4}$
$ \Rightarrow -1 \le x < \frac{1}{\sqrt{2}}$