Tangent of Sum of Three Angles/Proof 2

From ProofWiki
Jump to navigation Jump to search

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