How do you find the five remaining trigonometric function satisfying csctheta=-3/2, costheta<0?

1 Answer
Nghi N.
Feb 1, 2017

csc t = 1/(sin t) = - 3/2 --> sin t = - 2/3
sin t = - 2/3
cos^2 t = 1 - sin^2 t = 1 - 4/9 = 5/9 --> cos t = +- sqrt5/3
cos t = - sqrt5/3 (cos t < 0)
tan t = sin/(cos) = (- 2/3)(- 3/sqrt5) = 2/sqrt5 = (2sqrt5)/5
cot t = 1/(tan t) = 5/(2sqrt5).
sec t = 1/(cos) = - 3/sqrt5 = - (3sqrt5)/5.
csc t = -3/2

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