If `A+B+C=180^@ `, then `(cot A +cotB +cotC)/(cot A cotBcotC)` is equal to

Options

A. 1
B. `cot A cos B cot C`
C. `-1`
D. 0

Answer

To solve the problem, we need to find the value of \(\frac{\cot A + \cot B + \cot C}{\cot A \cot B \cot C}\) given that \(A + B + C = 180^\circ\).

### Step-by-Step Solution:

1. **Use the identity for cotangent of a sum**:
Since \(A + B + C = 180^\circ\), we can express \(C\) as \(C = 180^\circ - (A + B)\).
Using the cotangent identity:
\[ \cot(A + B + C) = \cot 180^\circ = 0 \]
We can write:
\[ \cot(A + B + C) = \frac{\cot A \cot B \cot C - (\cot A + \cot B + \cot C)}{\cot A \cot B + \cot B \cot C + \cot C \cot A} = 0 \]

2. **Set up the equation**:
From the identity, since the numerator must equal zero:
\[ \cot A \cot B \cot C - (\cot A + \cot B + \cot C) = 0 \]
This implies:
\[ \cot A + \cot B + \cot C = \cot A \cot B \cot C \]

3. **Substitute into the original expression**:
Now we can substitute this result into our original expression:
\[ \frac{\cot A + \cot B + \cot C}{\cot A \cot B \cot C} = \frac{\cot A \cot B \cot C}{\cot A \cot B \cot C} \]

4. **Simplify the expression**:
This simplifies to:
\[ \frac{\cot A \cot B \cot C}{\cot A \cot B \cot C} = 1 \]

5. **Final result**:
Therefore, the value of \(\frac{\cot A + \cot B + \cot C}{\cot A \cot B \cot C}\) is:
\[ 1 \]

### Conclusion:
Thus, the answer is \(1\).