In `DeltaABC, a^2+c^2=2002b^2` then `(cotA+cotC)/(cotB)` is equal to

Options

A. `(1)/(2001)`
B. `(2)/(2001)`
C. `(3)/(2001)`
D. `(4)/(2001)`

Answer

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

### Step-by-Step Solution:

1. **Use the Cotangent Definition**:
We know that \(\cot A = \frac{\cos A}{\sin A}\), \(\cot B = \frac{\cos B}{\sin B}\), and \(\cot C = \frac{\cos C}{\sin C}\). Therefore, we can express \((\cot A + \cot C) / \cot B\) as:
\[ \frac{\cot A + \cot C}{\cot B} = \frac{\frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}} = \frac{\cos A \sin B + \cos C \sin B}{\sin A \cos B + \sin C \cos B} \]

2. **Apply the Law of Cosines**:
Using the Law of Cosines, we can express \(\cos A\), \(\cos B\), and \(\cos C\):
- \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
- \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}\)
- \(\cos C = \frac{a^2 + b^2 - c^2}{2ab}\)

3. **Substitute Values**:
Now we will substitute the values of \(\cos A\), \(\cos B\), and \(\cos C\) into our expression. We also know from the problem statement that \(a^2 + c^2 = 2002b^2\).

4. **Simplify the Expression**:
After substituting and simplifying the expression, we will have:
\[ \frac{\frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + b^2 - c^2}{2ab}}{\frac{a^2 + c^2 - b^2}{2ac}} \]

5. **Use the Given Condition**:
Since \(a^2 + c^2 = 2002b^2\), we can replace \(a^2 + c^2\) in our expression. This will help us simplify further.

6. **Final Calculation**:
After performing the necessary algebraic manipulations and cancellations, we will arrive at:
\[ \frac{2b^2}{2001b^2} = \frac{2}{2001} \]

### Final Answer:
Thus, the value of \((\cot A + \cot C) / \cot B\) is:
\[ \frac{2}{2001} \]