In a triangle $\mathrm{ABC}$, with usual notations $\angle A=60^{\circ}$, then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

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Aicharya · Accepted Answer

$\begin{aligned} & \left(1+\frac{a}{c}+\frac{b}{c}ight)\left(1+\frac{c}{b}-\frac{a}{b}ight)=\frac{a+b+c}{c} 	imes \frac{b+c-a}{b}=\frac{(b+c)^2-a^2}{b c} \ & =\frac{b^2+c^2-a^2+2 b c}{b c}=2\left(\frac{b^2+c^2-a^2}{2 b c}ight)+2 \ & =2 \cos A+2=2 \cos 60^{\circ}+2=2 	imes \frac{1}{2}+2=3\end{aligned}$

In a triangle $\mathrm{ABC}$, with usual notations $\angle A=60^{\circ}$, then…

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