In triangle ABC, `(b-c)sin A+(c-a) sin B+( a-b) sin C=`

Options

A. `ab+bc+ca`
B. `a^2+b^2+c^2`
C. `0`
D. None of these

Answer

To solve the problem, we need to evaluate the expression \((b - c) \sin A + (c - a) \sin B + (a - b) \sin C\) in triangle ABC. We will use the sine rule and some algebraic manipulation to simplify the expression.

### Step-by-Step Solution:

1. **Use the Sine Rule**:
The sine rule states that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
From this, we can express \(a\), \(b\), and \(c\) in terms of \(k\):
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

2. **Substitute Values into the Expression**:
Substitute \(a\), \(b\), and \(c\) into the original expression:
\[ (b - c) \sin A + (c - a) \sin B + (a - b) \sin C \]
becomes:
\[ (k \sin B - k \sin C) \sin A + (k \sin C - k \sin A) \sin B + (k \sin A - k \sin B) \sin C \]

3. **Factor Out \(k\)**:
Factor out \(k\) from the expression:
\[ k \left[ (\sin B - \sin C) \sin A + (\sin C - \sin A) \sin B + (\sin A - \sin B) \sin C \right] \]

4. **Simplify the Expression**:
Now, we will simplify the expression inside the brackets:
\[ = k \left[ \sin B \sin A - \sin C \sin A + \sin C \sin B - \sin A \sin B + \sin A \sin C - \sin B \sin C \right] \]
Rearranging gives:
\[ = k \left[ \sin B \sin A - \sin B \sin C + \sin C \sin B - \sin C \sin A + \sin A \sin C - \sin A \sin B \right] \]

5. **Observe Cancellation**:
Notice that each term cancels out:
\[ \sin B \sin A - \sin B \sin C + \sin C \sin B - \sin C \sin A + \sin A \sin C - \sin A \sin B = 0 \]

6. **Final Result**:
Since the expression inside the brackets simplifies to zero:
\[ k \cdot 0 = 0 \]
Therefore, the final result is:
\[ (b - c) \sin A + (c - a) \sin B + (a - b) \sin C = 0 \]

### Conclusion:
The value of the expression \((b - c) \sin A + (c - a) \sin B + (a - b) \sin C\) is equal to **0**.