If `"cos"^(-1)x> "sin"^(-1)x`, then the set of all values of x is

Options

A. [0,1]
B. `[0,1/sqrt2)`
C. `[-1,1/sqrt2)`
D. None of these

Answer

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 identity, 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
Now, substituting this into our inequality gives:
\[ \frac{\pi}{2} - \sin^{-1} x > \sin^{-1} x \]

### Step 3: Rearrange the inequality
Rearranging the inequality, we have:
\[ \frac{\pi}{2} > 2 \sin^{-1} x \]
This simplifies to:
\[ \sin^{-1} x < \frac{\pi}{4} \]

### Step 4: Apply the sine function
Now we apply the sine function to both sides. Since the sine function is increasing in the interval \( [0, \frac{\pi}{2}] \), we can write:
\[ x < \sin\left(\frac{\pi}{4}\right) \]

### Step 5: Calculate the sine value
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ x < \frac{1}{\sqrt{2}} \]

### Step 6: Consider the domain of \( x \)
Since \( \cos^{-1} x \) and \( \sin^{-1} x \) are defined for \( x \) in the interval \( [-1, 1] \), we must also consider this constraint. Therefore, the final solution is:
\[ x \in [-1, \frac{1}{\sqrt{2}}) \]

### Final Answer
The set of all values of \( x \) is:
\[ [-1, \frac{1}{\sqrt{2}}) \]