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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(22)$