Tangent of Sum of Three Angles/Proof 2
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Theorem
- $\map \tan {A + B + C} = \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}$
Proof
\(\ds \map \tan {A + B + C}\) | \(=\) | \(\ds \dfrac {\tan A + \map \tan {B + C} } {1 - \tan A \tan {B + C} }\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A + \frac {\tan B + \tan C} {1 - \tan B \tan C} } {1 - \tan A \frac {\tan B + \tan C} {1 - \tan B \tan C} }\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A \paren {1 - \tan B \tan C} + \tan B + \tan C} {\paren {1 - \tan B \tan C} - \tan A \paren {\tan B + \tan C} }\) | multiplying top and bottom by $1 - \tan B \tan C$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}\) | simplification |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Tangents of sum and difference