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In any $\triangle A B C, \frac{1+\cos (A-B) \cdot \cos C}{1+\cos (A-C) \cdot \cos B}$ is equal to
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$\frac{a^2+b^2}{a^2+c^2}$
$\begin{aligned} & \frac{1+\cos (A-B) \cdot \cos \left(180^{\circ}-(A+B)\right)}{1+\cos (A-C) \cdot \cos \left(180^{\circ}-(A+C)\right)} \\ & =\frac{1-\cos (A-B) \cos (A+B)}{1-\cos (A-C) \cdot \cos (A+C)}\end{aligned}$
$\begin{aligned} & =\frac{1-\left(\cos ^2 A-\sin ^2 B\right)}{1-\left(\cos ^2 A-\sin ^2 C\right)}=\frac{1-\cos ^2 A+\sin ^2 B}{1-\cos ^2 A+\sin ^2 C} \\ & =\frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A+\sin ^2 C}=\frac{\frac{a^2}{k^2}+\frac{b^2}{k^2}}{\frac{a^2}{k^2}+\frac{c^2}{k^2}}=\frac{a^2+b^2}{a^2+c^2} \\ & \qquad\left(\begin{array}{c}\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ \therefore \sin A=a / k, \sin B=b / k, \\ \sin C=c / k\end{array}\right)\end{aligned}$
$\begin{aligned} & =\frac{1-\left(\cos ^2 A-\sin ^2 B\right)}{1-\left(\cos ^2 A-\sin ^2 C\right)}=\frac{1-\cos ^2 A+\sin ^2 B}{1-\cos ^2 A+\sin ^2 C} \\ & =\frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A+\sin ^2 C}=\frac{\frac{a^2}{k^2}+\frac{b^2}{k^2}}{\frac{a^2}{k^2}+\frac{c^2}{k^2}}=\frac{a^2+b^2}{a^2+c^2} \\ & \qquad\left(\begin{array}{c}\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ \therefore \sin A=a / k, \sin B=b / k, \\ \sin C=c / k\end{array}\right)\end{aligned}$
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