# Question #761c0

Dec 24, 2015

$x \in \pi \mathbb{Z}$

#### Explanation:

We know that $\tan \left(a + b\right) = \frac{\tan \left(a\right) + \tan \left(b\right)}{1 - \tan \left(a\right) \tan \left(b\right)}$ and $\tan \left(a - b\right) = \frac{\tan \left(a\right) - \tan \left(b\right)}{1 + \tan \left(a\right) \tan \left(b\right)}$. So we apply those formulas where they're needed.

$\tan \left(x\right) + \tan \left(120 + x\right) - \tan \left(120 - x\right) = 0 \iff \tan \left(x\right) + \frac{\tan \left(120\right) + \tan \left(x\right)}{1 - \tan \left(120\right) \tan \left(x\right)} - \frac{\tan \left(120\right) - \tan \left(x\right)}{1 + \tan \left(120\right) \tan \left(x\right)} = 0$.

Here, you switch the minus on the denominator of the 2nd fraction so you have the same denominator everywhere and everything gets canceled!

$\tan \left(x\right) + \tan \left(120 + x\right) - \tan \left(120 - x\right) = 0 \iff \tan \left(x\right) + \frac{\tan \left(120\right) + \tan \left(x\right) - \tan \left(120\right) - \tan \left(x\right)}{1 - \tan \left(120\right) \tan \left(x\right)} = 0 \iff \tan \left(x\right) = 0 \iff x \in \pi \mathbb{Z}$