In a triangle `ABC, 2 ac sin (1/2(A-B + C)) =`

Options

A. `a^(2) + b^(2) - c^(2)`
B. `c^(2) + a^(2) - b^(2)`
C. `b^(2) - c^(2) - a^(2)`
D. `c^(2) - a^(2) - b^(2)`

Answer

To solve the problem, we need to find the value of \( 2ac \sin\left(\frac{1}{2}(A - B + C)\right) \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Understand the Angles in Triangle**:
We know that the sum of angles in a triangle is:
\[ A + B + C = 180^\circ \]

2. **Express \( A + C \)**:
From the equation above, we can express \( A + C \) in terms of \( B \):
\[ A + C = 180^\circ - B \]

3. **Substituting into the Expression**:
We need to evaluate \( A - B + C \):
\[ A - B + C = (A + C) - B = (180^\circ - B) - B = 180^\circ - 2B \]

4. **Substituting into the Sine Function**:
Now we substitute this back into our original expression:
\[ 2ac \sin\left(\frac{1}{2}(A - B + C)\right) = 2ac \sin\left(\frac{1}{2}(180^\circ - 2B)\right) \]

5. **Using the Sine Identity**:
We can simplify \( \sin\left(\frac{1}{2}(180^\circ - 2B)\right) \):
\[ \sin\left(\frac{1}{2}(180^\circ - 2B)\right) = \sin(90^\circ - B) = \cos B \]

6. **Final Expression**:
Now substituting this back into our expression gives:
\[ 2ac \sin\left(\frac{1}{2}(A - B + C)\right) = 2ac \cos B \]

7. **Using the Cosine Rule**:
According to the cosine rule, we have:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

8. **Substituting the Cosine Value**:
Now substituting \( \cos B \) back into our expression:
\[ 2ac \cos B = 2ac \cdot \frac{a^2 + c^2 - b^2}{2ac} \]

9. **Simplifying**:
The \( 2ac \) terms cancel out:
\[ 2ac \cos B = a^2 + c^2 - b^2 \]

### Final Answer:
Thus, the value of \( 2ac \sin\left(\frac{1}{2}(A - B + C)\right) \) is:
\[ \boxed{a^2 + c^2 - b^2} \]