Triple Angle Formulas/Tangent/Proof 2

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Theorem

$\map \tan {3 \theta} = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$


Proof

\(\displaystyle \tan \left({3 \theta}\right)\) \(=\) \(\displaystyle \dfrac {\tan \theta + \tan \left({2 \theta}\right)} {1 - \tan \theta \tan \left({2 \theta}\right)}\) Tangent of Sum
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\tan \theta + \dfrac {2\tan \theta} {1 - \tan^2 \theta} } {1 - \tan \theta \dfrac {2 \tan \theta} {1 - \tan^2 \theta} }\) Double Angle Formula for Tangent
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\tan \theta \left({1 - \tan^2 \theta}\right) + 2 \tan \theta} {\left({1 - \tan^2 \theta}\right) - 2 \tan^2 \theta}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\tan \theta - \tan^3 \theta + 2 \tan \theta} {1 - \tan^2 \theta - 2 \tan^2 \theta}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}\) simplifying

$\blacksquare$