In triangle ABC,if frac{a+b+c}{BC+AB}+frac{a+ | vedclass

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(C) Given,$\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.Since $BC=a, AC=b, AB=c$,we have $\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.$\Rightarrow \frac{a+c+b}{a

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$\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

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(C) Given,$\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.Since $BC=a, AC=b, AB=c$,we have $\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.$\Rightarrow \frac{a+c+b}{a+c}+\frac{b+c+a}{b+c}=3$ $\Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3$.$\Rightarrow \frac{b}{a+c}+\frac{a}{b+c}=1$ $\Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)$.$\Rightarrow b^2+bc+a^2+ac=ab+ac+bc+c^2$.$\Rightarrow a^2+b^2-c^2=ab$.Dividing by $2ab$,we get $\frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}$.By the Cosine Rule,$\cos C = \frac{1}{2}$,so $C = 60^\circ = \frac{\pi}{3}$.We need to find $	an \frac{C}{8} = 	an \frac{\pi}{24}$.Using $	an \frac{	heta}{2} = \frac{1-\cos 	heta}{\sin 	heta}$,we have $	an \frac{\pi}{24} = \frac{1-\cos(\pi/12)}{\sin(\pi/12)}$.Since $\cos(\pi/12) = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin(\pi/12) = \frac{\sqrt{6}-\sqrt{2}}{4}$,$	an \frac{\pi}{24} = \frac{1-\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4-\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} 	imes \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}$.$= \frac{4\sqrt{6}+4\sqrt{2}-6-\sqrt{12}-\sqrt{12}-2}{6-2} = \frac{4\sqrt{6}+4\sqrt{2}-8-2\sqrt{3}}{4} = \sqrt{6}+\sqrt{2}-2-\frac{\sqrt{3}}{2}$.Wait,re-evaluating: $	an \frac{\pi}{24} = \sqrt{6}-\sqrt{3}+\sqrt{2}-2$.

In triangle $ABC$,if $\frac{a+b+c}{BC+AB}+\frac{a+b+c}{AC+AB}=3$,then $\tan \frac{C}{8}=$

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