A) \[\frac{1}{a+b+c}\]
B) \[\frac{2}{a+b+c}\]
C) \[\frac{3}{a+b+c}\]
D) None of these
Correct Answer: C
Solution :
\[\cos C=\frac{\pi }{3}\Rightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=ab\] Þ \[{{b}^{2}}+bc+{{a}^{2}}+ac=ab+ac+bc+{{c}^{2}}\] Þ \[b(b+c)+a(a+c)=(a+c)(b+c)\] Divide by \[(a+c)\text{ }(b+c)\] and add 2 on both sides Þ \[1+\frac{b}{a+c}+1+\frac{a}{b+c}=3\]Þ \[\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\].
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