A) 0
B) 1
C) \[\pm 1\]
D) 2
Correct Answer: A
Solution :
\[(b-c)\cot \frac{A}{2}=k(\sin B-\sin C)\cot \frac{A}{2}\] \[=2k\cos \frac{B+C}{2}\sin \frac{B-C}{2}\cot \frac{A}{2}\] \[=2k\sin \frac{A}{2}\,.\,\,\sin \frac{B-C}{2}\text{ }\text{. }\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[=2k\sin \left( \frac{B-C}{2} \right)\sin \left( \frac{B+C}{2} \right)=2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)\] or we get L.H.S. =\[\Sigma 2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)=0\].
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