If `alpha+beta-gamma=pi`, then `sin^(2)alpha+sin^(2)beta-sin^(2)gamma` is equal to

Options

A. `2sinalphasinbetacosgamma`
B. `2cosalphacosbetacosgamma`
C. `2sinalphasinbetasingamma`
D. None of these

Explanation

To solve the problem, we need to find the value of \( \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma \) given that \( \alpha + \beta - \gamma = \pi \).

### Step-by-Step Solution:

1. **Use the given equation**:
We start with the equation:
\[ \alpha + \beta - \gamma = \pi \]
Rearranging gives us:
\[ \gamma = \alpha + \beta - \pi \]

2. **Substitute for \(\sin^2 \gamma\)**:
We can express \(\sin^2 \gamma\) using the sine function:
\[ \sin \gamma = \sin(\alpha + \beta - \pi) = -\sin(\alpha + \beta) \]
Therefore:
\[ \sin^2 \gamma = \sin^2(\alpha + \beta - \pi) = \sin^2(\alpha + \beta) \]

3. **Use the sine addition formula**:
The sine addition formula states:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
Thus:
\[ \sin^2(\alpha + \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2 \]

4. **Expand \(\sin^2(\alpha + \beta)\)**:
Expanding this gives:
\[ \sin^2(\alpha + \beta) = \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta + \cos^2 \alpha \sin^2 \beta \]

5. **Combine the terms**:
Now substituting back into our expression:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta) \]

6. **Use the identity**:
We can use the identity \( \sin^2 A + \sin^2 B - \sin^2(A + B) = 2 \sin A \sin B \) for our case:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta) = 2 \sin \alpha \sin \beta \cos(\alpha - \beta) \]

7. **Final expression**:
Therefore, we conclude:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos(\alpha - \beta) \]

### Final Result:
Thus, the value of \( \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma \) is:
\[ \boxed{2 \sin \alpha \sin \beta \cos(\alpha - \beta)} \]