How do you evaluate $tan((5pi)/3)$ ?

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Answer 1 · 2015-04-26T02:21:02

The intersection of the ray for $(5pi)/3$ with the unit circle is at the same place as the ray for $(-pi/3)$

Again with reference to the unit circle that intersection point is at
$(x,y) = (1/2,-sqrt(3)/2)$

So
$tan((5pi)/3) = - sqrt(3)$

How do you evaluate tan((5pi)/3)tan((5pi)/3)?