# How do you prove tan(3theta)=(3tantheta-tan^3theta)/(1-3tan^2theta)?

Oct 7, 2016

see below

#### Explanation:

$\tan \left(3 \theta\right) = \frac{3 \tan \theta - {\tan}^{3} \theta}{1 - 3 {\tan}^{2} \theta}$

Left Side: $= \tan \left(3 \theta\right)$

$= \tan \left(2 \theta + \theta\right)$->Use the formulas $\tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan 2 A = \frac{2 \tan A}{1 - {\tan}^{2} A}$

$= \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta}$

$= \frac{\frac{2 \tan \theta}{1 - {\tan}^{2} \theta} + \tan \theta}{1 - \frac{2 \tan \theta}{1 - {\tan}^{2} \theta} \tan \theta}$

=((2tan theta+tantheta(1-tan^2theta))/(1-tan^2 theta))/((1-tan^2theta-2tan^2 theta)/(1-tan^2 theta)

$= \frac{2 \tan \theta + \tan \theta - {\tan}^{3} \theta}{1 - {\tan}^{2} \theta} \cdot \frac{1 - {\tan}^{2} \theta}{1 - {\tan}^{2} \theta - 2 {\tan}^{2} \theta}$

$= \frac{3 \tan \theta - {\tan}^{3} \theta}{\cancel{1 - {\tan}^{2} \theta}} \cdot \frac{\cancel{1 - {\tan}^{2} \theta}}{1 - 3 {\tan}^{2} \theta}$

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

$\therefore =$ Right Side