Question

Let θ be an angle in quadrant II, such that `tan θ = (-sqrt15)` how do you find cos θ and csc θ?

Answer 1

`cos t = -1/4`
`sin t = sqrt15/4`

Explanation:

Use the trig identity:
`cos^2 a = 1/(1 + tan^2 a)`
`cos^2 t = 1/(1 + 15) = 1/16` --> `cos t = +- 1/4`
Since t is in Quadrant II, then cos t is negative.
`cos t = - 1/4`
`sin t = tan t.cos t = (-sqrt15)(-1/4) = sqrt15/4`