正切分式定理

正切分式定理:
${\color{Red} \cfrac{\tan A}{\tan B}+ \cfrac{\tan A}{\tan C} =\cfrac{2a^2}{b^2+c^2-a^2} \qquad} $
${\color{Green} \cfrac{\tan B}{\tan A}+ \cfrac{\tan B}{\tan C} =\cfrac{2b^2}{a^2+c^2-b^2} \qquad} $
${\color{Violet} \cfrac{\tan C}{\tan A}+ \cfrac{\tan C}{\tan B} =\cfrac{2c^2}{a^2+b^2-c^2} \qquad } \cfrac{\tan C}{\tan A}+ \cfrac{\tan C}{\tan B} =\cfrac{2c^2}{a^2+b^2-c^2} \qquad $
$证明：\cfrac{\tan A}{\tan B}=\cfrac{\sin A\cos B}{\cos A \sin B}=\cfrac{a\times \cfrac{a^2+c^2-b^2}{2ac} }{b\times \cfrac{b^2+c^2-a^2}{2bc}}=\cfrac{a^2+c^2-b^2}{b^2+c^2-a^2} $
$\cfrac{\tan A}{\tan C}= \cfrac{\tan A}{\tan C}=\cfrac{\sin A \cos C}{\sin C \cos A}=\cfrac{a\times \cfrac{a^2+b^2-c^2}{2ab} }{c\times \cfrac{b^2+c^2-a^2}{2bc}}=\cfrac{a^2+b^2-c^2}{b^2+c^2-a^2} $
${\color{Red} 两式相加得证 } $

${\color{Red} 此题要用到正切恒等式及中线长定理 } $

${\color{Red} 此题用正切分式定理的证明过程或特殊值法，即分母相等 } $