In `DeltaABC`,
`(b-c)cotA/2+(c-a)cotB/2+(a-b)cotC/2=?`

Options

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

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{(b-c) \cot \frac{A}{2}}{1} + \frac{(c-a) \cot \frac{B}{2}}{1} + \frac{(a-b) \cot \frac{C}{2}}{1} \]

### Step 1: Use the cotangent half-angle formula
We know that:
\[ \cot \frac{A}{2} = \frac{s(s-a)}{\Delta}, \quad \cot \frac{B}{2} = \frac{s(s-b)}{\Delta}, \quad \cot \frac{C}{2} = \frac{s(s-c)}{\Delta} \]
where \(s\) is the semi-perimeter of the triangle \(ABC\) and \(\Delta\) is the area of triangle \(ABC\).

### Step 2: Substitute the cotangent values into the expression
Substituting these values into the expression, we get:
\[ (b-c) \frac{s(s-a)}{\Delta} + (c-a) \frac{s(s-b)}{\Delta} + (a-b) \frac{s(s-c)}{\Delta} \]

### Step 3: Factor out \(\frac{s}{\Delta}\)
Factoring out \(\frac{s}{\Delta}\) from the expression, we have:
\[ \frac{s}{\Delta} \left( (b-c)(s-a) + (c-a)(s-b) + (a-b)(s-c) \right) \]

### Step 4: Expand the terms inside the parentheses
Now, we will expand the terms:
1. \((b-c)(s-a) = bs - ba - cs + ca\)
2. \((c-a)(s-b) = cs - ca - bs + ab\)
3. \((a-b)(s-c) = as - ac - bs + bc\)

Combining these, we get:
\[ bs - ba - cs + ca + cs - ca - bs + ab + as - ac - bs + bc \]

### Step 5: Combine like terms
Now, we will combine like terms:
- The \(bs\) terms: \(bs - bs - bs = -bs\)
- The \(cs\) terms: \(-cs + cs = 0\)
- The \(as\) terms: \(as\)
- The \(ab\) terms: \(ab\)
- The \(ca\) terms: \(ca - ca = 0\)
- The \(ac\) terms: \(-ac\)
- The \(bc\) terms: \(bc\)

So, we have:
\[ -as + ab + bc - ac - bs \]

### Step 6: Simplify the expression
After simplification, we notice that all terms cancel out:
\[ 0 \]

### Final Result
Thus, the value of the expression is:
\[ \frac{s}{\Delta} \cdot 0 = 0 \]

### Conclusion
The final answer is:
\[ \boxed{0} \]