Question

Given the point (6,-8) on the terminal side of ø in standard position, find the exact value of csc ø?

Answer 1

`csc t = - 5/4`

Explanation:

Call t the arc (angle), that its terminal side passes at point (6, -8).
t is in Quadrant 4.
`tan t = y/x = -8/6 = - 4/3`
`cot = 1/(tan t) = - 3/4`
`sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 9/16) = 16/25`
`sin t = +- 4/5`
Since t is in Quadrant 4, then, sin t is negative
`sin t = - 4/5`
`csc t = 1/(sin t) = - 5/4`