Sum of Tangents of Angles in Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

$\tan A + \tan B + \tan C = \tan A \tan B \tan C$


Proof

\(\ds \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}\) \(=\) \(\ds \map \tan {A + B + C}\) Tangent of Sum of Three Angles
\(\ds \) \(=\) \(\ds \tan 180 \degrees\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 0\) Tangent of Straight Angle
\(\ds \leadsto \ \ \) \(\ds \tan A + \tan B + \tan C - \tan A \tan B \tan C\) \(=\) \(\ds 0\) multiplying both sides by $1 - \tan B \tan C - \tan C \tan A - \tan A \tan B$
\(\ds \leadsto \ \ \) \(\ds \tan A + \tan B + \tan C\) \(=\) \(\ds \tan A \tan B \tan C\)

$\blacksquare$


Sources