A) \[(b+c)\sin \frac{B-C}{2}=2a\cos \frac{A}{2}\]
B) \[(b+c)\cos \frac{A}{2}=2a\sin \frac{B-C}{2}\]
C) \[(b-c)\cos \frac{A}{2}=a\sin \frac{B-C}{2}\]
D) \[(b-c)\sin \frac{B-C}{2}=2a\cos \frac{A}{2}\]
Correct Answer: C
Solution :
\[\frac{b-c}{a}=\frac{\sin B-\sin C}{\sin A}=\frac{2\sin \frac{B-C}{2}\cos \frac{B+C}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}}=\frac{\sin \frac{B-C}{2}}{\cos \frac{A}{2}}\] Þ \[(b-c)\cos \frac{A}{2}=a\sin \frac{B-C}{2}\].
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