Standard way of using sine and cosine to find tangent

There was question I came upon, and I was stumped. The question was: evaluate the sine ,cosine, and tangent of the angle without using a calculator.

I was given $-\pi/6$. I know that sine is $-1/2$ and cosine is $\sqrt{3}/2$.

Normally I know that $\tan = \sin/\cos$. And doing so gives $-1/\sqrt{3}$. The problem was that the actual answer is $-\sqrt{3}/3$, and I can't seem to figure out how that can be possible

• $-\frac 1{\sqrt 3} = -\frac {\sqrt 3}{3}$. Those are both the same answer. – fleablood Nov 29 '17 at 7:09

$-\frac 1{\sqrt{3}} = -\frac 1{\sqrt{3}} * \frac {\sqrt{3}}{\sqrt{3}} = - \frac {\sqrt 3}{3}$.
$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt 3}$. To see this, rationalise the denominator of the latter by multiplying top and bottom by $\sqrt 3$.