If `cos^(-1)x gt sin^(-1) x`, then

Options

A. `1/(sqrt(2)) lt x le 1`
B. `0 le x le 1/(sqrt(2))`
C. `-1 le x lt 1/(sqrt(2))`
D. `x gt 0`

Explanation

To solve the inequality \( \cos^{-1} x > \sin^{-1} x \), we can follow these steps:

### Step 1: Use the identity
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
From this, we can express \( \cos^{-1} x \) in terms of \( \sin^{-1} x \):
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]

### Step 2: Substitute into the inequality
Substituting this into the inequality gives us:
\[ \frac{\pi}{2} - \sin^{-1} x > \sin^{-1} x \]

### Step 3: Rearrange the inequality
Now, we can rearrange this inequality:
\[ \frac{\pi}{2} > \sin^{-1} x + \sin^{-1} x \]
This simplifies to:
\[ \frac{\pi}{2} > 2 \sin^{-1} x \]

### Step 4: Divide by 2
Dividing both sides by 2 results in:
\[ \frac{\pi}{4} > \sin^{-1} x \]
or equivalently:
\[ \sin^{-1} x < \frac{\pi}{4} \]

### Step 5: Apply the sine function
Now, we can apply the sine function to both sides. Since the sine function is increasing in the range of \( \sin^{-1} x \):
\[ x < \sin\left(\frac{\pi}{4}\right) \]
Knowing that \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we can write:
\[ x < \frac{1}{\sqrt{2}} \]

### Step 6: Consider the range of \( \sin^{-1} x \)
The range of \( \sin^{-1} x \) is:
\[ -\frac{\pi}{2} \leq \sin^{-1} x \leq \frac{\pi}{2} \]
This implies:
\[ -1 \leq x \leq 1 \]

### Step 7: Combine the inequalities
Now we combine the inequalities:
1. From \( \sin^{-1} x < \frac{\pi}{4} \), we have \( x < \frac{1}{\sqrt{2}} \).
2. From the range of \( \sin^{-1} x \), we have \( -1 \leq x \leq 1 \).

Thus, we can conclude:
\[ -1 \leq x < \frac{1}{\sqrt{2}} \]

### Final Answer
The solution to the inequality \( \cos^{-1} x > \sin^{-1} x \) is:
\[ -1 \leq x < \frac{1}{\sqrt{2}} \]