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