Law of Tangents/Corollary/Proof 2

Corollary to Law of Tangents

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

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

$=$

$\ds \dfrac {2 R \sin A - 2 R \sin B} {2 R \sin A + 2 R \sin B}$

$\ds $

$=$

$\ds \dfrac {2 \cos \frac {A + B} 2 \sin \frac {A - B} 2} {2 \sin \frac {A + B} 2 \cos \frac {A - B} 2}$

$\ds $

$=$

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

$\ds $

$=$

$\ds \dfrac {\tan \frac {A - B} 2} {\tan \frac {180 \degrees - C} 2}$

$\ds $

$=$

$\ds \dfrac {\tan \frac {A - B} 2} {\map \tan {90^\circ - \frac C 2} }$

simplification

$\ds $

$=$

$\ds \dfrac {\tan \frac {A - B} 2} {\cot \dfrac C 2}$

$\ds \leadsto \ \ $

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

$=$

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

Hence the result.

$\blacksquare$