If `f(x)=sin^(-1)(sin x),g(x)=cos^(-1)(cos x)` and `h(x)=cot^(-1)(cot x)`, then which of the following is/are correct ?

Options

A. (a) `f(x)=g(x)=h(x)AA x in((pi)/(4),(pi)/(3))`
B. (b) `f(x)lt g(x)lt h(x)AA x in ((pi)/(2),pi)`
C. (c) `h(x)gt g(x)gt f(x)AA x in((3pi)/(2),2pi)`
D. (d) `f(x)gt g(x)` has no real solution

Explanation

To solve the problem, we need to analyze the functions \( f(x) = \sin^{-1}(\sin x) \), \( g(x) = \cos^{-1}(\cos x) \), and \( h(x) = \cot^{-1}(\cot x) \) over specified intervals. We will determine the relationships between these functions in the given intervals.

### Step 1: Understand the functions

1. **Function \( f(x) = \sin^{-1}(\sin x) \)**:
- The range of \( f(x) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
- It is periodic with a period of \( 2\pi \).
- For \( x \in [0, \pi] \), \( f(x) = x \).
- For \( x \in [\pi, 2\pi] \), \( f(x) = 2\pi - x \).

2. **Function \( g(x) = \cos^{-1}(\cos x) \)**:
- The range of \( g(x) \) is \( [0, \pi] \).
- It is periodic with a period of \( 2\pi \).
- For \( x \in [0, \pi] \), \( g(x) = x \).
- For \( x \in [\pi, 2\pi] \), \( g(x) = 2\pi - x \).

3. **Function \( h(x) = \cot^{-1}(\cot x) \)**:
- The range of \( h(x) \) is \( (0, \pi) \).
- It is periodic with a period of \( \pi \).
- For \( x \in (0, \pi) \), \( h(x) = x \).
- For \( x \in (\pi, 2\pi) \), \( h(x) = x - \pi \).

### Step 2: Analyze the intervals

1. **Interval \( \left[\frac{\pi}{4}, \frac{\pi}{3}\right] \)**:
- In this interval, \( f(x) \), \( g(x) \), and \( h(x) \) are all increasing.
- Therefore, \( f(x) = g(x) = h(x) \) holds true.

2. **Interval \( \left[\frac{\pi}{2}, \pi\right] \)**:
- Here, \( f(x) \) is decreasing, \( g(x) \) is increasing, and \( h(x) \) is also decreasing.
- Thus, \( f(x) < g(x) \) does not hold true.

3. **Interval \( \left[\frac{3\pi}{2}, 2\pi\right] \)**:
- In this interval, \( f(x) \) is decreasing, \( g(x) \) is decreasing, and \( h(x) \) is increasing.
- The relationship \( h(x) > g(x) > f(x) \) holds true.

4. **Interval \( (0, \pi) \)**:
- Here, \( f(x) \) is increasing, \( g(x) \) is increasing, and \( h(x) \) is increasing.
- Therefore, there is no real solution for \( f(x) > g(x) \).

### Conclusion

Based on the analysis, we can conclude the following:

- **Option A**: Correct, as \( f(x) = g(x) = h(x) \) for \( x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right] \).
- **Option B**: Incorrect, as \( f(x) < g(x) \) does not hold.
- **Option C**: Correct, as \( h(x) > g(x) > f(x) \) for \( x \in \left[\frac{3\pi}{2}, 2\pi\right] \).
- **Option D**: Correct, as there is no real solution for \( f(x) > g(x) \) in \( (0, \pi) \).