Tangent of Sum

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Theorem

$\map \tan {a + b} = \dfrac {\tan a + \tan b} {1 - \tan a \tan b}$

where $\tan$ is tangent.


Corollary

$\map \tan {a - b} = \dfrac {\tan a - \tan b} {1 + \tan a \tan b}$


Proof

\(\ds \map \tan {a + b}\) \(=\) \(\ds \frac {\map \sin {a + b} } {\map \cos {a + b} }\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {\sin a \cos b + \cos a \sin b} {\cos a \cos b - \sin a \sin b}\) Sine of Sum and Cosine of Sum
\(\ds \) \(=\) \(\ds \frac {\frac {\sin a} {\cos a} + \frac {\sin b} {\cos b} } {1 - \frac {\sin a \sin b} {\cos a \cos b} }\) dividing top and bottom by $\cos a \cos b$
\(\ds \) \(=\) \(\ds \frac {\tan a + \tan b} {1 - \tan a \tan b}\) Tangent is Sine divided by Cosine

$\blacksquare$


Sources

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