A) \[2\cos \frac{A-C}{2}\]
B) \[\sin \frac{A+C}{2}\]
C) \[\sin \frac{A}{2}\]
D) None of these
Correct Answer: A
Solution :
\[A+C=2B\Rightarrow B={{60}^{o}}\], \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] Since \[B={{60}^{o}}\]\[\Rightarrow ac={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\] Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\] Therefore \[\frac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}=\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B}\] \[=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{2\sin \frac{B}{2}\sin \frac{A+C}{2}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}}\] \[=\frac{\cos \frac{A-C}{2}}{\sin {{30}^{o}}}\Rightarrow 2\cos \frac{A-C}{2}\].
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