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In $\triangle A B C$, if $B+C=72^{\circ}$, then $\left(1+\frac{a}{c}+\frac{b}{c}\right)$ $\left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to
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$\frac{5-\sqrt{5}}{2}$
$\begin{aligned} & \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=\left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right) \\ & =\frac{(b+c)^2-a^2}{b c}=\frac{\left(b^2+c^2-a^2\right)+2 b c}{b c}\end{aligned}$
$\begin{array}{lr}=\frac{b^2+c^2-a^2}{b c}+2=2\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 & \\ =2 \cos A+2 & \left(\because B+C=72^{\circ}, \text { given }\right) \\ =2 \cos 108^{\circ}+2 & \quad\left(A+B+C=180^{\circ}\right) \\ =2 \cos \left(90^{\circ}+18^{\circ}\right)+2 & \quad\left[\Rightarrow A=180^{\circ}-(B+C]\right. \\ =-2 \sin 18^{\circ}+2 & \left(A=108^{\circ}\right] \\ =-2\left(\frac{\sqrt{5}-1}{4}\right)+2=\frac{1-\sqrt{5}+4}{2}=\frac{5-\sqrt{5}}{2}\end{array}$
$\begin{array}{lr}=\frac{b^2+c^2-a^2}{b c}+2=2\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 & \\ =2 \cos A+2 & \left(\because B+C=72^{\circ}, \text { given }\right) \\ =2 \cos 108^{\circ}+2 & \quad\left(A+B+C=180^{\circ}\right) \\ =2 \cos \left(90^{\circ}+18^{\circ}\right)+2 & \quad\left[\Rightarrow A=180^{\circ}-(B+C]\right. \\ =-2 \sin 18^{\circ}+2 & \left(A=108^{\circ}\right] \\ =-2\left(\frac{\sqrt{5}-1}{4}\right)+2=\frac{1-\sqrt{5}+4}{2}=\frac{5-\sqrt{5}}{2}\end{array}$
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