A) \[3\]
B) \[6\]
C) \[9\]
D) \[12\]
Correct Answer: B
Solution :
[b] \[\frac{\cos \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}=\frac{2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right)}{2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A+B}{2} \right)}\] \[=\frac{\sin A}{\sin C}+\frac{\sin B}{\sin C}\] \[\therefore \,\,\frac{\cos \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}+\frac{\cos \left( \frac{B-C}{2} \right)}{\cos \left( \frac{B+C}{2} \right)}+\frac{\cos \left( \frac{C-A}{2} \right)}{\cos \left( \frac{C+A}{2} \right)}\] \[=\left( \frac{\sin A}{\sin C}+\frac{\sin B}{\sin C} \right)+\left( \frac{\sin B}{\sin A}+\frac{\sin C}{\sin A} \right)+\left( \frac{\sin C}{\sin B}+\frac{\sin A}{\sin B} \right)\]\[=\left( \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} \right)+\left( \frac{\sin B}{\sin C}+\frac{\sin C}{\sin B} \right)+\left( \frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \right)\ge 6\]\[\left\{ \because \,\,for\,t>0,\,t+\frac{1}{t}\ge 2 \right\}\]You need to login to perform this action.
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