If `A+B+C=0,` then the value of `sum cot (B+C-A) cot (C+A-B)` is equal to

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of the expression:

\[ \sum \cot(B + C - A) \cot(C + A - B) \]

Given that \( A + B + C = 0 \), we can express \( B + C \) and \( C + A \) in terms of \( A \):

1. Since \( A + B + C = 0 \), we can rearrange this to find:
\[ B + C = -A \quad \text{and} \quad C + A = -B \]

2. Substitute these values into the expression:
\[ \sum \cot(B + C - A) \cot(C + A - B) = \sum \cot(-A - A) \cot(-B - B) \]
This simplifies to:
\[ \sum \cot(-2A) \cot(-2B) \]

3. Using the identity \( \cot(-\theta) = -\cot(\theta) \), we can rewrite this as:
\[ \sum (-\cot(2A)) (-\cot(2B)) = \sum \cot(2A) \cot(2B) \]

4. The expression now becomes:
\[ \cot(2A) \cot(2B) + \cot(2B) \cot(2C) + \cot(2C) \cot(2A) \]

5. We can factor out \( \cot(2B) \):
\[ \cot(2B) (\cot(2A) + \cot(2C)) + \cot(2C) \cot(2A) \]

6. Now, we need to express \( \cot(2A) + \cot(2C) \) in a manageable form. Using the identity for the cotangent of a sum:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
However, we notice that \( A + C = -B \), which leads to:
\[ \cot(2A + 2C) = \cot(-2B) = -\cot(2B) \]

7. Thus, we have:
\[ \cot(2A) + \cot(2C) = -\cot(2B) \]

8. Substitute this back into the expression:
\[ \cot(2B)(-\cot(2B)) + \cot(2C) \cot(2A) = -\cot^2(2B) + \cot(2C) \cot(2A) \]

9. Now, we can conclude that the sum simplifies to:
\[ 0 \]

Thus, the final answer is:

\[ \boxed{1} \]