How do you determine the exact values of the six trig function of the angle given (-4,10)?

On the trig unit circle,

tan a = x/y = 4/10 = 0.40 -> a = 21.80 deg

x = a + 90 = 90 + 21.80 = 111.80 deg (Quadrant II)

sin x = sin 111.80 = 0.93
cos x = cos 111.80 = - 0.37

`tan x = 0.93/-0.37 = - 2.50`
`cot x = 1/-2.50 = - 0.40`
`sec = 1/-0.37 = - 2.70`
`csc x = 1/0.93 = 1.07`

If point `(x,y)` is on the terminal side of an angle `theta` in standard position, then the trigonometric functions are defined using `r` which is the distance between `(x,y)` and the origin: `r = sqrt(x^2+y^2)`
In this case `r = sqrt((-4)^2 + (10)^2) = sqrt 116 = sqrt (4*29) = 2sqrt29`

`sin theta = y/r` In this case `sin theta = 10/(2sqrt29) = 5/sqrt29 = (5sqrt29)/29`

`cos theta = x/r`, in this case `cos theta = -4/(2sqrt29) = - 2/sqrt29 = -(2sqrt29)/29`

`tan theta = y/x`. In this case `tan theta = 10/(-4) = -5/2`

the other functions are the reciprocals of these:

`csc theta = r/y = (2sqrt29)/10 = sqrt29/5`

`sec theta = r/x = (2sqrt29)/-4 = -sqrt29/2`

`cot theta = x/y = (-4)/10 = -2/5`.

It is very important to memorize the definitions of the trigonometric functions. (Flash card are a great idea -- make some.)