Let `f(x) = min ( tan^(-1) x, cot^(-1)x) " and " h ( x) = f ( x + 2) - pi //3 ". Let " x_(1), x_(2) ( "where " x_(1) lt x_(2))` be the integers in the range of h (x) , then the value of ` ( cos^(-1) ( cos x_(1)) + sin ^(-1) ( sin x_2)) ` is equal to

Answer

To solve the problem step-by-step, we will follow these instructions:

### Step 1: Define the functions
We have:
- \( f(x) = \min(\tan^{-1} x, \cot^{-1} x) \)
- \( h(x) = f(x + 2) - \frac{\pi}{3} \)

### Step 2: Analyze the function \( f(x) \)
To find \( f(x) \), we need to determine where \( \tan^{-1} x \) and \( \cot^{-1} x \) intersect.

1. **Finding the intersection**:
Set \( \tan^{-1} x = \cot^{-1} x \).
Recall that \( \cot^{-1} x = \tan^{-1} \left(\frac{1}{x}\right) \).
Therefore, we have:
\[ \tan^{-1} x = \tan^{-1} \left(\frac{1}{x}\right) \]
This implies:
\[ x = \frac{1}{x} \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \]

2. **Determine the behavior of \( f(x) \)**:
- For \( x < -1 \): \( f(x) = \cot^{-1} x \)
- For \( -1 < x < 1 \): \( f(x) = \tan^{-1} x \)
- For \( x > 1 \): \( f(x) = \cot^{-1} x \)

### Step 3: Analyze the function \( h(x) \)
Now we need to evaluate \( h(x) \):
\[ h(x) = f(x + 2) - \frac{\pi}{3} \]

1. **Determine the intervals for \( x + 2 \)**:
- For \( x + 2 < -1 \) (i.e., \( x < -3 \)): \( f(x + 2) = \cot^{-1}(x + 2) \)
- For \( -1 < x + 2 < 1 \) (i.e., \( -3 < x < -1 \)): \( f(x + 2) = \tan^{-1}(x + 2) \)
- For \( x + 2 > 1 \) (i.e., \( x > -1 \)): \( f(x + 2) = \cot^{-1}(x + 2) \)

### Step 4: Find the range of \( h(x) \)
1. **Evaluate \( h(x) \) for different intervals**:
- **For \( x < -3 \)**:
\[ h(x) = \cot^{-1}(x + 2) - \frac{\pi}{3} \]
- **For \( -3 < x < -1 \)**:
\[ h(x) = \tan^{-1}(x + 2) - \frac{\pi}{3} \]
- **For \( x > -1 \)**:
\[ h(x) = \cot^{-1}(x + 2) - \frac{\pi}{3} \]

2. **Find the minimum and maximum values**:
- As \( x \to -3 \), \( h(-3) = \cot^{-1}(-1) - \frac{\pi}{3} = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12} \).
- As \( x \to -1 \), \( h(-1) = \tan^{-1}(1) - \frac{\pi}{3} = \frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12} \).

Thus, the range of \( h(x) \) is from \( -\frac{\pi}{12} \) to \( \frac{5\pi}{12} \).

### Step 5: Identify integers in the range
The integers \( x_1 \) and \( x_2 \) in the range of \( h(x) \) are:
- \( x_1 = -2 \)
- \( x_2 = -1 \)

### Step 6: Calculate the final expression
We need to evaluate:
\[ \cos^{-1}(\cos x_1) + \sin^{-1}(\sin x_2 \]
Substituting \( x_1 = -2 \) and \( x_2 = -1 \):
\[ \cos^{-1}(\cos(-2)) + \sin^{-1}(\sin(-1)) \]

1. **Evaluate \( \cos^{-1}(\cos(-2)) \)**:
Since \( -2 \) is in the range of \( [0, \pi] \):
\[ \cos^{-1}(\cos(-2)) = 2 \]

2. **Evaluate \( \sin^{-1}(\sin(-1)) \)**:
Since \( -1 \) is in the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \):
\[ \sin^{-1}(\sin(-1)) = -1 \]

### Final Calculation:
\[ \cos^{-1}(\cos(-2)) + \sin^{-1}(\sin(-1)) = 2 - 1 = 1 \]

### Final Answer:
The value is \( \boxed{1} \).

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