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

$$\begin{gathered} {\cos }^{-1}x>{\sin }^{-1}x \\ \Rightarrow {\cos }^{-1}x>\frac{\mathrm{\pi }}{2}-{\cos }^{-1}x \\ \Rightarrow 2{\cos }^{-1}x>\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\cos }^{-1}x>\frac{\mathrm{\pi }}{4} \\ \Rightarrow x>\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow x>\frac{1}{\sqrt{2}} \end{gathered}$$
We know that the maximum value of cosine fuction is 1.
$\therefore \frac{1}{\sqrt{2}}<x\le 1$
Hence, the correct answer is option(a).