Double Angle Formulas/Tangent/Proof 1
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Theorem
- $\tan 2 \theta = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$
Proof
\(\ds \tan 2 \theta\) | \(=\) | \(\ds \frac {\sin 2 \theta} {\cos 2 \theta}\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \cos \theta \sin \theta} {\cos^2 \theta - \sin^2 \theta}\) | Double Angle Formula for Sine and Double Angle Formula for Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\frac {2 \cos \theta \sin \theta} {\cos^2 \theta} } {\frac {\cos^2 \theta - \sin^2 \theta} {\cos^2 \theta} }\) | dividing numerator and denominator by $\cos^2 \theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\) | simplifying: Tangent is Sine divided by Cosine |
$\blacksquare$