已知 $BC^2 + AC^2 = 5AB^2$ , 求 $\frac{\tan C}{\tan A} + \frac{\tan C}{\tan B}$
解，因为 $BC^2 + AC^2 = 5AB^2$ 可得 $a^2 + b^2 = 5c^2$ ，由此可联想到余弦定理，因此
$$
\begin{flalign}
\frac{\tan C}{\tan A} + \frac{\tan C}{\tan B} \\
={\tan C}(\frac{\cos A}{sinA} + \frac{\cos B}{sinB}) \\
= (\frac{sinC}{\cos C})(\frac{\cos A}{sinA} + \frac{\cos B}{sinB}) \\
= \frac{sinC\cos A}{\cos C\sin A} + \frac{sinC\cos B}{\cos CsinB} \\
= \frac{c\cdot \cos A}{a\cdot \cos C} + \frac{c \cdot \cos B}{b \cdot \cos C} \\
= \frac{bc \cdot \cos A + ac \cdot \cos B}{ab \cdot \cos C} \\
= \frac{2bc \cdot \cos A + 2ac \cdot \cos B}{2ab \cdot \cos C} \\
= \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2}{a^2 + b^2 - c^2} \\
= \frac{2c^2}{4c^2} \\
= \frac{1}{2}
\end{flalign}
$$
其中第 4 步 到 5 步 引入下面的正弦定理的推论
$$
\begin{align}
\frac{a}{sinA} = \frac{b}{sinB} = \frac{b}{sinC} \\
=> \frac{sinC}{sinA} = \frac{c}{a} \\
=> \frac{sinC}{sinB} = \frac{c}{b}
\end{align}
$$
第 7 步 到第 8 步，引入了下面的余弦定理的推论
$$
\begin{align}
a^2 = b^2 + c^2 - 2bc\cdot cosA => 2bc \cdot cosA = b^2 + c^2 - a^2 \\
b^2 = a^2 + c^2 - 2ac \cdot cosB => 2ac \cdot cosB = a^2 + c^2 - b^2 \\
c^2 = a^2 + b^2 - 2ab \cdot cosC => 2ab \cdot cosC = a^2 + b^2 - c^2
\end{align}
$$
