# How do you write in terms of theta tan(theta-pi/6)? Thanks.

Mar 1, 2018

$\textcolor{b l u e}{\tan \left(\theta - \frac{\pi}{6}\right) = \frac{\sqrt{3} \tan \theta - 1}{\sqrt{3} + \tan \theta}}$

#### Explanation:

So we have
$\tan \left(\theta - \frac{\pi}{6}\right)$

color (green)(Tan (A-B)=(tan A-tanB)/(1+tan A tanB)

Here $A = \theta \text{ ,} B = \frac{\pi}{6}$

Plugging in values

$\tan \left(\theta - \frac{\pi}{6}\right) = \frac{\tan \theta - \tan \left(\frac{\pi}{6}\right)}{1 + \tan \theta \cdot \tan \left(\frac{\pi}{6}\right)}$

$\textcolor{red}{\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}}$

$\tan \left(\theta - \frac{\pi}{6}\right) = \frac{\tan \theta - \frac{1}{\sqrt{3}}}{1 + \tan \theta \cdot \frac{1}{\sqrt{3}}}$

$\tan \left(\theta - \frac{\pi}{6}\right) = \frac{\left(\sqrt{3} \tan \theta - 1\right) \cdot \cancel{\frac{1}{\sqrt{3}}}}{\left(\sqrt{3} + \tan \theta\right) \cdot \cancel{\frac{1}{\sqrt{3}}}}$

$\tan \left(\theta - \frac{\pi}{6}\right) = \frac{\sqrt{3} \tan \theta - 1}{\sqrt{3} + \tan \theta}$

Hope it helps!