How do you find the exact value of the six trigonometric functions of the angle whose terminal side passes through #(-10/3, -2/3)#?

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Answer 1 · 2018-02-27T10:26:22

As detailed below.

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#r = sqrt(x^2 + y^2) = sqrt((-10/3)^2 + (-2/3)^2) ) = sqrt(104/9)#

#sin theta = y / r =(-2/3) / sqrt(104/9) = -1/sqrt26#

#csc theta = 1/sin theta = -sqrt26#

#cos theta = x / r = (-10/3) / sqrt(104/9 )= -5 / sqrt 26#

#sec theta = 1 / cos theta = -sqrt26 / 5#

#tan theta = y / x = -(2/3) / -(10/3) = 1/ 5#

#cot theta = 1 / tan theta = 5#