In a `DeltaA B Csum(b+c)tanA/2tan((B-C)/2)=`

Options

A. a
B. b
C. c
D. 0

Answer

To solve the problem given in the triangle \( \Delta ABC \), we need to evaluate the expression:

\[ \sum (b+c) \tan\left(\frac{A}{2}\right) \tan\left(\frac{B-C}{2}\right) \]

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (b+c) \tan\left(\frac{A}{2}\right) \tan\left(\frac{B-C}{2}\right) \). Here, \( A, B, C \) are the angles of the triangle opposite to the sides \( a, b, c \) respectively.

2. **Using the Identity for \( \tan\left(\frac{B-C}{2}\right)**: We can rewrite \( \tan\left(\frac{B-C}{2}\right) \) using the identity:
\[ \tan\left(\frac{B-C}{2}\right) = \frac{\tan\left(\frac{B}{2}\right) - \tan\left(\frac{C}{2}\right)}{1 + \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right)} \]
However, for our purpose, we can also express it as:
\[ \tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot\left(\frac{A}{2}\right) \]

3. **Substituting Back into the Expression**:
Substitute this back into the expression:
\[ (b+c) \tan\left(\frac{A}{2}\right) \cdot \frac{b-c}{b+c} \cot\left(\frac{A}{2}\right) \]
Here, \( (b+c) \) cancels out:
\[ = (b-c) \tan\left(\frac{A}{2}\right) \cot\left(\frac{A}{2}\right) \]

4. **Using the Identity \( \tan x \cot x = 1 \)**:
Since \( \tan\left(\frac{A}{2}\right) \cot\left(\frac{A}{2}\right) = 1 \):
\[ = (b-c) \]

5. **Repeating for Other Angles**:
We repeat this process for the other angles \( B \) and \( C \):
- For \( B \): \( (c-a) \)
- For \( C \): \( (a-b) \)

6. **Summing Up**:
Now we sum these results:
\[ (b-c) + (c-a) + (a-b) \]

7. **Simplifying the Sum**:
When we simplify this:
\[ = b - c + c - a + a - b = 0 \]

### Final Answer:
Thus, the final answer is:
\[ \sum (b+c) \tan\left(\frac{A}{2}\right) \tan\left(\frac{B-C}{2}\right) = 0 \]