# How is $-(\pi - \tan^{-1}(1/\sqrt{3}))=-5 \pi/6$

How did they get from 2nd last to last step?

$$-\Bigg(\pi - \tan^{-1}\bigg({\frac{1}{\sqrt{3}}}\bigg)\Bigg)=-\frac{5 \pi}{6}$$

-
It's $\arctan\frac1{\sqrt 3}$, not $\arctan\frac13$. –  Ｊ. Ｍ. Nov 21 '11 at 10:50
There is a mistake: $\frac{\sqrt{5}}{\sqrt{15}} = \frac{1}{\sqrt 3}$. After you fix it, you see that $\tan^{-1}\frac{1}{\sqrt 3} = \frac{\pi}{6}$.
...and if you don't know why $\tan^{-1}\frac{1}{\sqrt 3} = \frac{\pi}{6}$, you'll want to remember the 30-60-90 triangle... –  Ｊ. Ｍ. Nov 21 '11 at 10:54