Question

How do you use the point (-9,14) on the terminal side of the angle to evaluate the six trigonometric functions?

Answer 1

The terminal side, containing point (-9, 14), is in Quadrant 3.
Call t the angle.
`tan t = -14/9`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 196/81) = 81/277`
`cos t = +- 9/sqrt277`
Because t is in Quadrant 3, cos t is negative -->
`cos t = - 9/(sqrt277)`
`sin t = tan t.cos t = (-14/9)(-9/sqrt277) = 14/sqrt277`
`tan t = -14/9`
`cot t = - 9/14`
`sec t = 1/(cos) = - sqrt277/9`
`csc t = 1/(sin) = sqrt277/14`