In a triangle ABC, if a, b and c are the lengths of the sides opposite to the angles A, B and C respectively, then what is \( \frac{\sin(A-B)}{\sin(A+B)} \) equal to ?
- A. \frac{a^2}{a^2-b^2}
- B. \frac{a^2}{a^2+b^2}
- C. \frac{a^2-b^2}{c^2} ✓
- D. \frac{a^2-b^2}{b^2}
Correct Answer: C. \frac{a^2-b^2}{c^2}
Explanation
Multiply numerator and denominator by \sin(A+B). \frac{\sin(A-B)\sin(A+B)}{\sin^2(A+B)} = \frac{\sin^2 A - \sin^2 B}{\sin^2 C}. Using the sine rule, this converts to \frac{k^2 a^2 - k^2 b^2}{k^2 c^2} = \frac{a^2 - b^2}{c^2}.