Solve this


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

(a) $\frac{1}{\sqrt{2}}

(b) $0 \leq x<\frac{1}{\sqrt{2}}$

(c) $-1 \leq x<\frac{1}{\sqrt{2}}$

(d) $x>0$


$\cos ^{-1} x>\sin ^{-1} x$

$\Rightarrow \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x$

$\Rightarrow 2 \cos ^{-1} x>\frac{\pi}{2}$

$\Rightarrow \cos ^{-1} x>\frac{\pi}{4}$

$\Rightarrow x>\cos \frac{\pi}{4}$

$\Rightarrow x>\frac{1}{\sqrt{2}}$

We know that the maximum value of cosine fuction is 1 .

$\therefore \frac{1}{\sqrt{2}}

Hence, the correct answer is option(a).

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