Step 1: Understanding the Question:
This is an inequality involving inverse trigonometric functions. The domain of both functions is \( [-1, 1] \).
Step 2: Key Formula or Approach:
1. Factorize the expression: \( (\cos^{-1} x - \sin^{-1} x)(\cos^{-1} x + \sin^{-1} x)>0 \).
2. Use the identity: \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).
3. Since \( \frac{\pi}{2}>0 \), the inequality simplifies to \( \cos^{-1} x - \sin^{-1} x>0 \implies \cos^{-1} x>\sin^{-1} x \).
Step 3: Detailed Explanation:
We need \( \cos^{-1} x>\sin^{-1} x \) for \( x \in [-1, 1] \).
At \( x = \frac{1}{\sqrt{2}} \), \( \cos^{-1} x = \sin^{-1} x = \frac{\pi}{4} \).
For \( x<\frac{1}{\sqrt{2}} \), \( \cos^{-1} x \) increases and \( \sin^{-1} x \) decreases (graphically).
For example, at \( x = 0 \), \( \cos^{-1}(0) = \frac{\pi}{2} \) and \( \sin^{-1}(0) = 0 \), which satisfies \( \frac{\pi}{2}>0 \).
At \( x = -1 \), \( \cos^{-1}(-1) = \pi \) and \( \sin^{-1}(-1) = -\frac{\pi}{2} \), which satisfies \( \pi>-\frac{\pi}{2} \).
Thus, the condition holds for \( x \in [-1, \frac{1}{\sqrt{2}}) \).
Step 4: Final Answer:
The solution is \( -1 \le x<\frac{1}{\sqrt{2}} \).