Question

Given Tan t = -4 and csc t > 0, how do you find the other 5 trig function?

Answer 1

Call tan x = t. Use the trig identity: `1 + t^2 = 1/(cos^2 x)`
Since tan x is negative and csc x > 0, the arc x must be in Quadrant II of the trig unit circle

`cos^2 x = 1/(1 + 16) = 1/17 --> cos x = - 24` (Quadrant II)

`sin ^2 x = 1 - cos^2 x = 1 - 0.06 = 0.94 -> sin x = 0.97` (Quadrant II)

`cot x = -0.24/0.97 = 0.25`

`sec x = -4.16`

$csc x = 1.03#