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In any $\triangle A B C, \frac{b-c \cos A}{c-b \cos A}$ is equal to
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$\frac{\cos C}{\cos B}$
$\begin{aligned} \frac{b-c \cos A}{c-b \cos A} & =\frac{b-\frac{b^2+c^2-a^2}{2 b}}{c-\frac{b^2+c^2-a^2}{2 c}} \\ & =\left(\frac{b^2+a^2-c^2}{c^2+a^2-b^2}\right) \times \frac{c}{b} \\ & =\frac{b^2+a^2-c^2}{2 a b} \cdot \frac{2 a c}{c^2+a^2-b^2} \\ & =\frac{\cos C}{\cos B}\end{aligned}$
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