How do you find the exact value of the five remaining trigonometric function given #csctheta=-5# and the angle is in standard position in quadrant IV?

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Answer 1 · 2016-11-19T18:18:44

Please see the explanation.

Given: #csc(theta) = -5#

Use the identity #sin(theta) = 1/csc(theta)#:

#sin(theta) = -1/5#

Use the identity #cos(theta) = +-sqrt(1-sin^2(theta))# but in quadrant IV the cosine is positive so drop the negative:

#cos(theta) = sqrt(1-(-1/5)^2)#

#cos(theta) = (2sqrt(6))/5#

Use the identity #sec(theta) = 1/cos(theta)#:

#sec(theta) = 5/(2sqrt(6)) = (5sqrt(6))/12#

Use the identity #tan(theta) = sin(theta)/cos(theta)#:

#tan(theta) = (-1/5)/((2sqrt(6))/5) = -sqrt(6)/12#

Use the identity #cot(theta) = cos(theta)/sin(theta)#

#cot(theta) = ((2sqrt(6))/5)/(-1/5) = -2sqrt(6)#