In triangle `A B C ,2a csin(1/2(A-B+C))` is equal to `a^2+b^2-c^2` (b) `c^2+a^2-b^2` `b^2-c^2-a^2` (d) `c^2-a^2-b^2`

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)`

Explanation

To solve the problem, we need to show that in triangle \( ABC \), the expression \( 2a c \sin\left(\frac{1}{2}(A - B + C)\right) \) is equal to \( c^2 + a^2 - b^2 \).

### Step-by-Step Solution:

1. **Understand the Angles in Triangle**:
We know that in any triangle, the sum of the angles is \( 180^\circ \). Therefore, we can express \( A + B + C = 180^\circ \). From this, we can derive:
\[ A + C = 180^\circ - B \]

2. **Rearranging the Angles**:
We can rearrange the above equation to find:
\[ A + C - B = 180^\circ - 2B \]

3. **Using the Sine Function**:
We need to evaluate \( 2a c \sin\left(\frac{1}{2}(A - B + C)\right) \). We can substitute for \( A + C - B \):
\[ 2a c \sin\left(\frac{1}{2}(A - B + C)\right) = 2a c \sin\left(\frac{1}{2}(180^\circ - 2B)\right) \]

4. **Applying the Sine Identity**:
Using the sine identity \( \sin(90^\circ - x) = \cos(x) \), we have:
\[ \sin\left(\frac{1}{2}(180^\circ - 2B)\right) = \cos(B) \]
Therefore, we can rewrite our expression as:
\[ 2a c \cos(B) \]

5. **Using the Cosine Rule**:
According to the cosine rule, we know:
\[ \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting this into our expression gives:
\[ 2a c \cos(B) = 2a c \left(\frac{a^2 + c^2 - b^2}{2ac}\right) \]

6. **Simplifying the Expression**:
The \( 2ac \) in the numerator and denominator cancels out:
\[ 2a c \cdot \frac{a^2 + c^2 - b^2}{2ac} = a^2 + c^2 - b^2 \]

7. **Final Result**:
Thus, we have shown that:
\[ 2a c \sin\left(\frac{1}{2}(A - B + C)\right) = a^2 + c^2 - b^2 \]

### Conclusion:
The correct answer is \( c^2 + a^2 - b^2 \), which corresponds to option (b).