Law of Tangents

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The Law of Tangents is a rather obscure trigonometric identity that is sometimes used in place of its better-known counterparts, the law of sines and law of cosines, to calculate angles or sides in a triangle.


If $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then \[\frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} .\]


Let $s$ and $d$ denote $(A+B)/2$, $(A-B)/2$, respectively. By the Law of Sines, \[\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .\] By the angle addition identities, \[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]}\] as desired. $\square$



This problem has not been edited in. If you know this problem, please help us out by adding it.


In $\triangle ABC$, let $D$ be a point in $BC$ such that $AD$ bisects $\angle A$. Given that $AD=6,BD=4$, and $DC=3$, find $AB$.


Show that $[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$.

(AoPS Vol. 2)

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