Question

Question #92df2

Answer 1

`tan t = y/x = -5/sqrt3 = - (5sqrt3)/3`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 25/3) = 3/28`
`cos t = +- sqrt3/(2sqrt7)`
t is in Quadrant 2, therefor cos t is negative
`cos t = - sqrt3/(2sqrt7) = - sqrt21/14`
`sin^2 t = 1 - cos^2 t = 1 - 3/28 = 25/28`
`sin t = 5/(2sqrt7) = (5sqrt7)/14` (sin t > 0, since t is in Q. 2)
`cot t = 1/tan t = - 3/(5sqrt3) = - sqrt3/5`.
`sec t = 1/(cos t) = -14/sqrt21 = - (14sqrt21)/21 = (2sqrt21)/3`
`csc t = 1/(sin t) = 14/(5sqrt7) = (14sqrt7)/35`