The value of `cot^(-1)(cot(-10))+cos^(-1)(cos10)` is given by `a +bpi`, a, b being rational numbers, then (b-a) equals……

Answer

To solve the problem, we need to evaluate the expression \( \cot^{-1}(\cot(-10)) + \cos^{-1}(\cos(10)) \) and express it in the form \( a + b\pi \), where \( a \) and \( b \) are rational numbers. Finally, we will find \( b - a \).

### Step-by-Step Solution:

1. **Evaluate \( \cot^{-1}(\cot(-10)) \)**:
- We know that \( \cot(-\theta) = -\cot(\theta) \). Therefore, \( \cot(-10) = -\cot(10) \).
- Using the property of inverse cotangent, \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \), we have:
\[ \cot^{-1}(\cot(-10)) = \cot^{-1}(-\cot(10)) = \pi - \cot^{-1}(\cot(10)) \]

2. **Determine the range of \( \cot^{-1}(x) \)**:
- The function \( \cot^{-1}(x) \) is defined such that \( \cot^{-1}(x) = x \) for \( x \) in the range \( (0, \pi) \).
- Since \( 10 \) is outside this range, we need to adjust it. We can express \( 10 \) as \( 4\pi - (4\pi - 10) \).

3. **Rewrite \( \cot^{-1}(\cot(10)) \)**:
- We can express \( 10 \) as \( 4\pi - (4\pi - 10) \):
\[ \cot^{-1}(\cot(10)) = \cot^{-1}(\cot(4\pi - (4\pi - 10))) = \cot^{-1}(\cot(4\pi - 10)) \]

4. **Evaluate \( \cos^{-1}(\cos(10)) \)**:
- Similarly, we can express \( \cos(10) \) as \( \cos(4\pi - (4\pi - 10)) \).
- Since \( \cos^{-1}(x) \) is defined for \( x \) in the range \( [0, \pi] \), we have:
\[ \cos^{-1}(\cos(10)) = 4\pi - 10 \]

5. **Combine the results**:
- Now we can combine both parts:
\[ \cot^{-1}(\cot(-10)) + \cos^{-1}(\cos(10)) = (\pi - \cot^{-1}(\cot(4\pi - 10))) + (4\pi - 10) \]
- Since \( \cot^{-1}(\cot(4\pi - 10)) = 4\pi - 10 \) (as \( 4\pi - 10 \) is in the range \( (0, \pi) \)):
\[ = \pi - (4\pi - 10) + (4\pi - 10) \]
- This simplifies to:
\[ = \pi + 10 - 4\pi + 4\pi - 10 = 8\pi - 20 \]

6. **Identify \( a \) and \( b \)**:
- We can express this as:
\[ 8\pi - 20 = -20 + 8\pi \]
- Thus, \( a = -20 \) and \( b = 8 \).

7. **Calculate \( b - a \)**:
- Finally, we find:
\[ b - a = 8 - (-20) = 8 + 20 = 28 \]

### Final Answer:
The value of \( b - a \) is \( 28 \).