Question

How do you find all six trigonometric function of `theta` if the point `(-9a, -12a)` is on the terminal side of `theta`?

Answer 1

cos t = - 3/5
sin t = - 4/5
cot t = 3/4
sec t = - 5/3
csct = - 5/4

Explanation:

t is in Quadrant 3.
`tan t = y/x = -12a/-9a = 4/3`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 16/9) = 9/25`
`cos t = +- 3/5`
Since t is in Quadrant 3, take the negative answer.
`sin^2 t = 1 - cos^2 t = 1 - 9/25 = 16/25`.
`sin t = +- 4/5`
Take the negative answer, because t is in Q. 3.
`cot t = 3/4`
`sec t = 1/(cos t) = - 5/3`
`csc t = 1/(sin t) = - 5/4`