In triangle ABC,if angle C = frac{pi}{3},then | vedclass

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(B) Given $\angle C = \frac{\pi}{3}$.Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $\frac{

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$\frac{1}{b+c}$

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(B) Given $\angle C = \frac{\pi}{3}$.Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$,which implies $a^2+b^2-c^2 = ab$,or $a^2+b^2 = ab+c^2$.Adding $ac+bc$ to both sides,we get $a^2+b^2+ac+bc = ab+c^2+ac+bc$.Factoring both sides,$a(a+c) + b(b+c) = (a+b)(c+c)$ is not the path; rather $a(a+c) + b(b+c) = (a+b)(c+c)$ is incorrect. Let's re-factor: $a^2+ac+b^2+bc = ab+c^2+ac+bc \Rightarrow a(a+c) + b(b+c) = (a+b)(c+c)$ is wrong. Correct factorization: $a^2+b^2-ab = c^2$.Adding $ac+bc$ to both sides: $a^2+b^2-ab+ac+bc = c^2+ac+bc \Rightarrow a(a+c) + b(b+c) - ab = c(a+b+c)$.Actually,the standard identity is: $\frac{a}{b+c} + \frac{b}{a+c} = 1$.Adding $1$ to both sides: $\frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} = 3$.Dividing by $(a+b+c)$: $\frac{1}{b+c} + \frac{1}{a+c} = \frac{3}{a+b+c}$.Therefore,$\frac{3}{a+b+c} - \frac{1}{a+c} = \frac{1}{b+c}$.

In $\triangle ABC$,if $\angle C = \frac{\pi}{3}$,then $\frac{3}{a+b+c} - \frac{1}{a+c}$ equals

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