In `DeltaABC`,
`(b-c)cot(A/2)+(c-a)cot(B/2)+(a-b)cot(C/2)=?`

Options

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

Answer

To solve the equation \((b-c)\cot\left(\frac{A}{2}\right) + (c-a)\cot\left(\frac{B}{2}\right) + (a-b)\cot\left(\frac{C}{2}\right) = ?\) for triangle \(ABC\), we will follow these steps:

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

### Step 2: Substitute the cotangent values
Substituting these identities into the original equation gives:
\[ (b-c)\frac{s(s-a)}{K} + (c-a)\frac{s(s-b)}{K} + (a-b)\frac{s(s-c)}{K} \]

### Step 3: Factor out common terms
Factoring out \(\frac{s}{K}\) from the equation:
\[ \frac{s}{K} \left[ (b-c)(s-a) + (c-a)(s-b) + (a-b)(s-c) \right] \]

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

Combining these:
\[ bs - ba - cs + ca + cs - cb - as + ab + as - ac - bs + ba \]

### Step 5: Simplify the expression
Notice that the terms will cancel out:
- \(bs\) and \(-bs\)
- \(-ba\) and \(ba\)
- \(-cs\) and \(cs\)
- \(-as\) and \(as\)
- \(-cb\) and \(ab\)
- \(-ac\) and \(ca\)

Thus, we are left with:
\[ 0 \]

### Conclusion
Therefore, the final result is:
\[ (b-c)\cot\left(\frac{A}{2}\right) + (c-a)\cot\left(\frac{B}{2}\right) + (a-b)\cot\left(\frac{C}{2}\right) = 0 \]