if `A+B+C=pi` then `sin^(2) A + sin^(2) B - sin^(2) C =`

Options

A. 2 sin A sin B sin C
B. 2cos A cos B cosC
C. 2cos A sin B sin C
D. 2 sin A sin B cos C

Answer

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

### Step-by-Step Solution:

1. **Use the identity for \( C \)**:
Since \( A + B + C = \pi \), we can express \( C \) as:
\[ C = \pi - (A + B) \]

2. **Apply the sine identity**:
Using the sine identity, we have:
\[ \sin(\pi - x) = \sin x \]
Therefore, we can say:
\[ \sin C = \sin(\pi - (A + B)) = \sin(A + B) \]

3. **Expand \( \sin(A + B) \)**:
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

4. **Square \( \sin C \)**:
Now, we square \( \sin C \):
\[ \sin^2 C = \sin^2(A + B) = (\sin A \cos B + \cos A \sin B)^2 \]
Expanding this, we get:
\[ \sin^2 C = \sin^2 A \cos^2 B + 2 \sin A \cos A \sin B \cos B + \cos^2 A \sin^2 B \]

5. **Substitute into the original expression**:
Now we substitute \( \sin^2 C \) back into the original expression:
\[ \sin^2 A + \sin^2 B - \sin^2 C = \sin^2 A + \sin^2 B - (\sin^2 A \cos^2 B + 2 \sin A \cos A \sin B \cos B + \cos^2 A \sin^2 B) \]

6. **Simplify the expression**:
This simplifies to:
\[ \sin^2 A + \sin^2 B - \sin^2 A \cos^2 B - 2 \sin A \cos A \sin B \cos B - \cos^2 A \sin^2 B \]
Rearranging the terms gives:
\[ \sin^2 A (1 - \cos^2 B) + \sin^2 B (1 - \cos^2 A) - 2 \sin A \cos A \sin B \cos B \]
Using \( 1 - \cos^2 x = \sin^2 x \):
\[ \sin^2 A \sin^2 B + \sin^2 B \sin^2 A - 2 \sin A \cos A \sin B \cos B \]
This simplifies to:
\[ \sin^2 A + \sin^2 B - \sin^2 C = \sin^2 A + \sin^2 B - \sin^2(A + B) \]

7. **Final Result**:
After simplification, we find:
\[ \sin^2 A + \sin^2 B - \sin^2 C = 1 \]

### Final Answer:
\[ \sin^2 A + \sin^2 B - \sin^2 C = 1 \]