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

$-1 \le x < \frac{1}{\sqrt{2}}$

The correct answer is Option (3) → $-1 \le x < \frac{1}{\sqrt{2}}$ ##

We know the identity:
cos⁻¹x + sin⁻¹x = π/2

Given:
cos⁻¹x > sin⁻¹x

Substitute:
π/2 − sin⁻¹x > sin⁻¹x
⇒ π/2 > 2 sin⁻¹x
⇒ sin⁻¹x < π/4

Now applying sine on both sides:
x < sin(π/4) = 1/√2

Also, domain of x ∈ [−1, 1]

So,
−1 ≤ x < 1/√2

Alternatively

We know that,

Range of $\cos^{-1} x = [0, \pi]$

and range of $\sin^{-1} x = \left[ \frac{-\pi}{2}, \frac{\pi}{2} \right]$

From the graph,

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

When $x \in \left[ -1, \frac{1}{\sqrt{2}} \right]$

$⇒-1 \le x < \frac{1}{\sqrt{2}}$