A) \[\sqrt{3}\]
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
\[\because \] \[3a=b+c\] \[\Rightarrow \] \[\frac{b+c}{a}=3\] Applying sine rule, we get \[\frac{\sin B+\sin C}{\sin A}=3\] \[\Rightarrow \] \[\frac{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}}=3\] \[\Rightarrow \] \[\frac{\cos \frac{A}{2}\cos \frac{B-C}{2}}{\cos \left( \frac{B+C}{2} \right)\cos \frac{A}{2}}=3\] \[\Rightarrow \] \[\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\sin \frac{C}{2}=\] \[3\left[ \cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2} \right]\] \[\Rightarrow \] \[2\cos \frac{B}{2}\cos \frac{C}{2}=4\sin \frac{B}{2}\sin \frac{C}{2}\] \[\Rightarrow \] \[\cot \frac{B}{2}\cot \frac{C}{2}=2\] Alternate Solution: We have, \[3a=b+c\] ?(i) Now, \[\cot \frac{B}{2}\cot \frac{C}{2}\] \[\Rightarrow \] \[\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}.\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}\] \[=\frac{s}{s-a}=\frac{\frac{a+b+c}{2}}{\frac{a+b+c}{2}-a}\] \[=\frac{a+b+c}{-a+b+c}\] \[=\frac{a+3a}{-a+3a}\] [from (i)] \[=\frac{4a}{2a}=2\]
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