Tangent of Sum of Three Angles/Proof 1

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

\(\ds \map \sin {A + B + C}\)

\(=\)

\(\ds \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C\)

Sine of Sum of Three Angles

\(\ds \map \cos {A + B + C}\)

\(=\)

\(\ds \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C\)

Cosine of Sum of Three Angles

\(\ds \leadsto \ \ \)

\(\ds \map \tan {A + B + C}\)

\(=\)

\(\ds \dfrac {\sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C} {\cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C}\)

Tangent is Sine divided by Cosine

\(\ds \)

\(=\)

\(\ds \dfrac {\frac {\sin A \cos B \cos C} {\cos A \cos B \cos C} + \frac {\cos A \sin B \cos C} {\cos A \cos B \cos C} + \frac {\cos A \cos B \sin C} {\cos A \cos B \cos C} - \frac {\sin A \sin B \sin C} {\cos A \cos B \cos C} } {\frac {\cos A \cos B \cos C} {\cos A \cos B \cos C} - \frac {\sin A \sin B \cos C} {\cos A \cos B \cos C} - \frac {\sin A \cos B \sin C} {\cos A \cos B \cos C} - \frac {\cos A \sin B \sin C} {\cos A \cos B \cos C} }\)

dividing numerator and denominator by $\cos A \cos B \cos C$

\(\ds \)

\(=\)

\(\ds \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}\)

Tangent is Sine divided by Cosine and simplifying

$\blacksquare$