Question

How do you find the values of the six trigonometric functions given `sectheta=-2` and `sintheta>0`?

Answer 1

`sec t = 1/(cos t) = - 2` --> `cos t = - 1/2`
Find sin t.
`sin^2 t = 1 - cos^2 t = 1 - 1/4 = 3/4` --> `sin t = +- sqrt3/2`
Since sin t > 0, t is in Quadrant II, there for:
`sin t = sqrt3/2`.
`tan t = sin/(cos) = (sqrt3/2)(- 2/1) = - sqrt3`
`cot t = 1/(tan) = - 1/sqrt3 = - sqrt3/3`
`csc t = 1/(sin) = 2/sqrt3 = (2sqrt3)/3`