Sum of Cotangents of Half Angles in Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

$\cot \dfrac A 2 + \cot \dfrac B 2 + \cot \dfrac C 2 = \cot \dfrac A 2 \cot \dfrac B 2 \cot \dfrac C 2$


Proof

\(\ds \dfrac {\cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2 - \cot \frac A 2 \cot \frac B 2 \cot \frac C 2} {1 - \cot \frac B 2 \cot \frac C 2 - \cot \frac C 2 \cot \frac A 2 - \cot \frac A 2 \cot \frac B 2}\) \(=\) \(\ds \map \cot {\frac A 2 + \frac B 2 + \frac C 2}\) Cotangent of Sum of Three Angles
\(\ds \) \(=\) \(\ds \cot 90 \degrees\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 0\) Cotangent of Right Angle
\(\ds \leadsto \ \ \) \(\ds \cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2 - \cot \frac A 2 \cot \frac B 2 \cot \frac C 2\) \(=\) \(\ds 0\) multiplying both sides by $1 - \cot \dfrac B 2 \cot \dfrac C 2 - \cot \dfrac C 2 \cot \dfrac A 2 - \cot \dfrac A 2 \cot \dfrac B 2$
\(\ds \leadsto \ \ \) \(\ds \cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2\) \(=\) \(\ds \cot \frac A 2 \cot \frac B 2 \cot \frac C 2\)

$\blacksquare$


Sources

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