How do you use fundamental identities to find the values of the trigonometric values given angle in quadrant III, such that sec(x) = -3?

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Answer 1 · 2016-12-23T18:53:12

Please see the explanation.

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

#cos(x) = -1/3#

Use the identity #sin(x) = +-sqrt(1 - cos^2(x))#

Because we are given that x is in the third quadrant, we know that the sine function is negative so we remove the #+#sign:

#sin(x) = -sqrt(1 - cos^2(x))#

Substitute #(-1/3)^2" for "cos^2(x)#

#sin(x) = -sqrt(1 - (-1/3)^2)#

#sin(x) = -sqrt(1 - 1/9)#

#sin(x) = -sqrt(8/9)#

#sin(x) = (-2sqrt(2))/3#

Use the identity #csc(x) = 1/sin(x)#

#csc(x) = (-3sqrt(2))/4#

Use the identity #tan(x) = sin(x)/cos(x)#

#tan(x) = ((-2sqrt(2))/3)/(-1/3) = 2sqrt(2)#

Use the identity #cot(x) = 1/tan(x)#:

#cot(x) = 1/(2sqrt(2)) = sqrt(2)/4#