`(b-c)cot""1/2A+(c-a)cot""1/2B+(a-b)cot""1/2C=……`

Answer

To solve the expression \((b-c)\cot\frac{1}{2}A + (c-a)\cot\frac{1}{2}B + (a-b)\cot\frac{1}{2}C\), we will use the formula for \(\cot\frac{1}{2}A\), \(\cot\frac{1}{2}B\), and \(\cot\frac{1}{2}C\) in terms of the sides of the triangle and its area.

### Step-by-Step Solution:

1. **Understanding the Semi-perimeter and Area**:
- Let \(s\) be the semi-perimeter of the triangle, defined as:
\[ s = \frac{a + b + c}{2} \]
- Let \(\Delta\) be the area of the triangle.

2. **Using the Cotangent Formula**:
- We know that:
\[ \cot\frac{1}{2}A = \frac{s(s-a)}{\Delta}, \quad \cot\frac{1}{2}B = \frac{s(s-b)}{\Delta}, \quad \cot\frac{1}{2}C = \frac{s(s-c)}{\Delta} \]

3. **Substituting the Cotangent Values**:
- Substitute these values into the expression:
\[ (b-c)\cot\frac{1}{2}A + (c-a)\cot\frac{1}{2}B + (a-b)\cot\frac{1}{2}C \]
- This becomes:
\[ (b-c)\frac{s(s-a)}{\Delta} + (c-a)\frac{s(s-b)}{\Delta} + (a-b)\frac{s(s-c)}{\Delta} \]

4. **Factoring Out Common Terms**:
- Factor out \(\frac{s}{\Delta}\):
\[ \frac{s}{\Delta} \left( (b-c)(s-a) + (c-a)(s-b) + (a-b)(s-c) \right) \]

5. **Expanding the Expression**:
- Expand the terms inside the parentheses:
\[ (b-c)(s-a) = bs - ba - cs + ca \]
\[ (c-a)(s-b) = cs - cb - as + ab \]
\[ (a-b)(s-c) = as - ac - bs + bc \]

6. **Combining Like Terms**:
- Combine all the expanded terms:
\[ (bs - ba - cs + ca) + (cs - cb - as + ab) + (as - ac - bs + bc) \]
- After combining, we notice that all terms cancel out:
- \(bs\) cancels with \(-bs\)
- \(-ba\) cancels with \(ab\)
- \(-cs\) cancels with \(cs\)
- \(-as\) cancels with \(as\)
- \(-cb\) cancels with \(bc\)
- \(-ac\) cancels with \(ca\)

7. **Final Result**:
- Since all terms cancel out, we have:
\[ \frac{s}{\Delta} \cdot 0 = 0 \]
- Therefore, the value of the expression is:
\[ 0 \]