Law of Tangents/Corollary/Proof 1

Corollary to Law of Tangents

$\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$

$\ds \dfrac {a + b} {a - b}$

$=$

$\ds \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }$

$\ds \leadsto \ \ $

$\ds \tan \frac {A - B} 2$

$=$

$\ds \dfrac {a - b} {a + b} \tan \frac {A + B} 2$

algebraic manipulation

$\ds $

$=$

$\ds \dfrac {a - b} {a + b} \tan \frac {180 \degrees - C} 2$

$\ds $

$=$

$\ds \dfrac {a - b} {a + b} \map \tan {90 \degrees - \frac C 2}$

$\ds $

$=$

$\ds \dfrac {a - b} {a + b} \cot \dfrac C 2$

$\blacksquare$