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, \[\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .\]

(In general, since $\frac{x}{sin X}$ is constant in a triangle, any ratio of linear combinations applied to lengths of sides is equal to the ratio of the same linear combinations applied to the sines of the angles of the same sides.)

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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