In any triangle `A B C ,sin^2A-sin^2B+sin^2C` is always equal to (A) `2sinAsinBcosC` (B) `2sinAcosBsinC` (C) `2sinAcosBcosC` (D) `2sinAsinBsinC`

Explanation

To solve the problem, we need to evaluate the expression \( \sin^2 A - \sin^2 B + \sin^2 C \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Use the Triangle Angle Sum Property**:
Since \( A + B + C = \pi \), we can express \( A + B \) as:
\[
A + B = \pi - C
...