In `Delta ABC, a ^(2) + c^(2)= 2002 b ^(2)` then `(cotB)/(cot A + cotC) ` is equal to `"________"`

Answer

To solve the problem, we need to find the value of \(\frac{\cot B}{\cot A + \cot C}\) given that \(a^2 + c^2 = 2002b^2\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Cotangent Formula**:
The cotangent of an angle in a triangle can be expressed in terms of the sides of the triangle. Specifically:
\[ \cot B = \frac{a^2 + c^2 - b^2}{4 \Delta} \]
\[ \cot A = \frac{b^2 + c^2 - a^2}{4 \Delta} \]
\[ \cot C = \frac{a^2 + b^2 - c^2}{4 \Delta} \]
where \(\Delta\) is the area of triangle \(ABC\).

2. **Substituting the Given Condition**:
We know from the problem statement that:
\[ a^2 + c^2 = 2002b^2 \]
We can substitute this into the cotangent formulas.

3. **Finding \(\cot B\)**:
Substitute \(a^2 + c^2\) into the formula for \(\cot B\):
\[ \cot B = \frac{2002b^2 - b^2}{4 \Delta} = \frac{2001b^2}{4 \Delta} \]

4. **Finding \(\cot A + \cot C\)**:
Now, we need to calculate \(\cot A + \cot C\):
\[ \cot A = \frac{b^2 + c^2 - a^2}{4 \Delta} \]
\[ \cot C = \frac{a^2 + b^2 - c^2}{4 \Delta} \]
Adding these two:
\[ \cot A + \cot C = \frac{(b^2 + c^2 - a^2) + (a^2 + b^2 - c^2)}{4 \Delta} = \frac{2b^2}{4 \Delta} = \frac{b^2}{2 \Delta} \]

5. **Calculating \(\frac{\cot B}{\cot A + \cot C}\)**:
Now we can substitute \(\cot B\) and \(\cot A + \cot C\) into the expression:
\[ \frac{\cot B}{\cot A + \cot C} = \frac{\frac{2001b^2}{4 \Delta}}{\frac{b^2}{2 \Delta}} = \frac{2001b^2}{4 \Delta} \cdot \frac{2 \Delta}{b^2} \]
Simplifying this gives:
\[ = \frac{2001 \cdot 2}{4} = \frac{4002}{4} = 1000.5 \]

### Final Answer:
Thus, the value of \(\frac{\cot B}{\cot A + \cot C}\) is \(1000.5\).