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 a triangle \( ABC \).

### Step-by-Step Solution:

1. **Use the identity for sine squared difference**:
We start with the expression:
\[
\sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B)
...