STATEMENT -1 : If `[sin^(-1)x] gt [cos^(-1)x]`,where [] represents the greatest integer function, then `x in [sin1,1]` is
and
STATEMENT -2 : `cos^(-1)(cosx)=x,x in [-1,1]`

Options

A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -2
B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -2
C. Statement-1 is True, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To solve the given problem, we need to analyze the two statements provided:

### Statement 1:
If \([ \sin^{-1} x ] > [ \cos^{-1} x ]\), where \([]\) represents the greatest integer function, then \(x \in [\sin(1), 1]\).

### Statement 2:
\(\cos^{-1}(\cos x) = x\), for \(x \in [-1, 1]\).

### Step-by-Step Solution:

#### Step 1: Analyze Statement 2
We start with Statement 2: \(\cos^{-1}(\cos x) = x\).

1. **Understanding the Definition**: The function \(\cos^{-1}(y)\) is defined for \(y \in [-1, 1]\) and its range is \([0, \pi]\).
2. **Condition for Equality**: The equality \(\cos^{-1}(\cos x) = x\) holds true if \(x\) is in the range of \(\cos^{-1}\), which means \(x\) must be in \([0, \pi]\).
3. **Conclusion for Statement 2**: Therefore, Statement 2 is true for \(x\) in the interval \([0, \pi]\) but not necessarily for all \(x \in [-1, 1]\). Thus, Statement 2 is **false** as stated.

#### Step 2: Analyze Statement 1
Now, let's analyze Statement 1: \([ \sin^{-1} x ] > [ \cos^{-1} x ]\).

1. **Understanding the Functions**:
- The function \(\sin^{-1}(x)\) is defined for \(x \in [-1, 1]\) and its range is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- The function \(\cos^{-1}(x)\) is defined for \(x \in [-1, 1]\) and its range is \([0, \pi]\).

2. **Greatest Integer Function**: The greatest integer function \([]\) takes the floor of the value.
- For \(x = 0\), \(\sin^{-1}(0) = 0\) and \(\cos^{-1}(0) = \frac{\pi}{2}\). Thus, \([ \sin^{-1}(0) ] = 0\) and \([ \cos^{-1}(0) ] = 1\).
- For \(x = 1\), \(\sin^{-1}(1) = \frac{\pi}{2}\) and \(\cos^{-1}(1) = 0\). Thus, \([ \sin^{-1}(1) ] = 1\) and \([ \cos^{-1}(1) ] = 0\).

3. **Finding the Intersection**:
- The two functions intersect at \(x = \frac{1}{\sqrt{2}}\) (approximately 0.707). For \(x > \frac{1}{\sqrt{2}}\), \(\sin^{-1}(x) > \cos^{-1}(x)\).
- Therefore, if \([ \sin^{-1} x ] > [ \cos^{-1} x ]\), it implies that \(x\) must be in the interval \((\frac{1}{\sqrt{2}}, 1]\).

4. **Conclusion for Statement 1**: Thus, Statement 1 is true as \(x\) must be in the interval \((\frac{1}{\sqrt{2}}, 1]\).

### Final Conclusion:
- **Statement 1** is true.
- **Statement 2** is false.

### Answer:
The correct option is **C**: Statement 1 is true and Statement 2 is false.