If `A+B+C=180^(@)`, then `Sigmatan.(A)/(2)tan.(B)/(2)` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the value of \( \sum \frac{\tan(A/2)}{\tan(B/2)} \) given that \( A + B + C = 180^\circ \).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We know that \( A + B + C = 180^\circ \). This implies that \( C = 180^\circ - (A + B) \).

2. **Using Half-Angle Identities**:
From the condition, we can express \( A/2 + B/2 = 90^\circ - C/2 \). This means:
\[ \frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2} \]

3. **Applying the Tangent Addition Formula**:
We can use the tangent addition formula:
\[ \tan\left(\frac{A}{2} + \frac{B}{2}\right) = \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right)}{1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right)} \]
Since \( \frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2} \), we have:
\[ \tan\left(90^\circ - \frac{C}{2}\right) = \cot\left(\frac{C}{2}\right) \]

4. **Setting Up the Equation**:
Therefore, we can equate:
\[ \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right)}{1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right)} = \cot\left(\frac{C}{2}\right) \]

5. **Expressing Cotangent**:
Recall that \( \cot\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{C}{2}\right)} \). Thus, we can rewrite the equation:
\[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) = \cot\left(\frac{C}{2}\right) \left(1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right)\right) \]

6. **Final Value**:
After simplifying, we find that:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) + \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) = 1 \]
This implies:
\[ \sum \frac{\tan(A/2)}{\tan(B/2)} = 1 \]

### Conclusion:
Thus, the value of \( \sum \frac{\tan(A/2)}{\tan(B/2)} \) is **1**.