# 正切定理

${\displaystyle {\frac {a-b}{a+b}}={\frac {\mathrm {tan} \,{\frac {\alpha -\beta }{2}}}{\mathrm {tan} \,{\frac {\alpha +\beta }{2}}}}}$
${\displaystyle {\frac {b-c}{b+c}}={\frac {\mathrm {tan} \,{\frac {\beta -\gamma }{2}}}{\mathrm {tan} \,{\frac {\beta +\gamma }{2}}}}}$
${\displaystyle {\frac {c-a}{c+a}}={\frac {\mathrm {tan} \,{\frac {\gamma -\alpha }{2}}}{\mathrm {tan} \,{\frac {\gamma +\alpha }{2}}}}}$

## 证明

${\displaystyle {\frac {a+b}{a-b}}}$开始，由正弦定理得出

{\displaystyle {\begin{aligned}{\frac {a+b}{a-b}}&={\frac {a{\frac {\sin \alpha }{a}}+b{\frac {\sin \beta }{b}}}{a{\frac {\sin \alpha }{a}}-b{\frac {\sin \beta }{b}}}}\\&={\frac {\sin \alpha +\sin \beta }{\sin \alpha -\sin \beta }}\\&={\frac {2\sin[{\frac {1}{2}}(\alpha +\beta )]\cos[{\frac {1}{2}}(\alpha -\beta )]}{2\cos[{\frac {1}{2}}(\alpha +\beta )]\sin[{\frac {1}{2}}(\alpha -\beta )]}}\end{aligned}}}
（和差化积，参阅三角恒等式
{\displaystyle {\begin{aligned}{\frac {a+b}{a-b}}&={\frac {\sin[{\frac {1}{2}}(\alpha +\beta )]\cos[{\frac {1}{2}}(\alpha -\beta )]}{\cos[{\frac {1}{2}}(\alpha +\beta )]\sin[{\frac {1}{2}}(\alpha -\beta )]}}\\&={\frac {\tan[{\frac {1}{2}}(\alpha +\beta )]}{\tan[{\frac {1}{2}}(\alpha -\beta )]}}\end{aligned}}}

## 参考资料

1. ^ See Eli Maor, Trigonometric Delights, Princeton University Press, 2002.