Question

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

Answer 1

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)`