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

1 Answer
May 18, 2018

#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#