How do you find the 6 trigonometric functions for 600 degrees?

First thing, let's see how many full loops we can get out of that angle, `theta = 600^"o" = 360^"o" + 240^"o" = 2*360^"o" - 120^"o"`

Why this is important, you may ask? It's because every 360^"o" degrees we have a full loop and return to where we were. So for the two main trig functions, `sin(x)` and `cos(x)` we can say that

`sin(600^"o") = sin(240^"o") = sin(-120^"o")`
`cos(600^"o") = cos(240^"o") = cos(-120^"o")`

And since every other function is a ratio of one or two of these functions, we can work with that. Now, we know that
`sin(-theta) = -sin(theta)` and `cos(-theta) = cos(theta)`

So we can say that
`sin(600^"o") = -sin(120^"o")`
`cos(600^"o") = cos(120^"o")`

We can then use the angle sum / double angle formula, since `120^"o" = 60^"o"+60^"o"`
`-sin(120^"o") = -(sin(60^"o")cos(60^"o") + sin(60^"o")cos(60^"o"))`
`cos(120^"o") = cos(60^"o")cos(60^"o") - sin(60^"o")sin(60^"o"))`

Since `60^"o"` is a special angle we know that `sin(60^"o") = sqrt(3)/2` and `cos(60^"o") = 1/2`, so

`-sin(120^"o") = -(sqrt(3)/2*1/2 + sqrt(3)/2*1/2) = -sqrt(3)/2`
`cos(120^"o") = 1/2*1/2- sqrt(3)/2*sqrt(3)/2 = -2/4 = -1/2`

And from that, we just evaluate the others by division:
`sec(600^"o") = 1/cos(600^"o") = -1/(1/2) =-2`
`csc(600^"o") = 1/sin(600^"o") = -1/(sqrt(3)/2) = -2/sqrt(3) = -(2sqrt(3))/3`
`tan(600^"o") = sin(600^"o")/cos(600^"o") = (sqrt(3)/2)/(1/2) = sqrt3`
`cot(600^"o") = 1/tan(600^"o") = 1/(sqrt(3)) = sqrt(3)/3`

600 = 240 + 360 deg
`sin 600 = sin 240 = - sin 60 = - sqrt3/2`
`cos 600 = cos 240 = - cos 60 = - 1/2`
`tan 600 = tan 240 = (-sqrt3/2)/-(1/2) = sqrt3` (arc 240 is in Quadrant III, its tan is positive)
`cot 600 = 1/(tan) = 1/sqrt3 = sqrt3/3`
`sec 600 = 1/(cos) = - 2`
`csc 600 = 1/(sin) = - (2sqrt3)/3`