How do you find all six trigonometric function of #theta# if the point #(sqrt3,-1)# is on the terminal side of #theta#?

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Answer 1 · 2017-08-25T17:43:07

Given: #(x,y) = (sqrt3,-1)#

Compute r:

#r = sqrt(x^2+y^2)

#r = sqrt((sqrt3)^2+(-1)^2)#

#r = sqrt(4)#

#r = 2#

#sin(theta) = y/r#

#sin(theta) = -1/2" [1]"#

#cos(theta) = x/r#

#cos(theta) = sqrt3/2" [2]"#

#csc(theta) = 1/sin(theta)#

#csc(theta) = -2" [3]"#

#sec(theta) = 1/cos(theta)#

#sec(theta) = (2sqrt3)/3" [4]"#

#tan(theta) = sin(theta)sec(theta)#

#tan(theta) = -sqrt3/3" [5]"#

#cot(theta) = 1/tan(theta)#

#cot(theta) = -sqrt3" [6]"#