What are the exact values of the six trigonometric functions for - 7pi/6 radians?

For - 7pi/6 is an angle in second quadrant, then sine and cosecant must be positive; and cosine, secant, tangent and cotangent must me negative.

The reference angle is:

7pi/6-pi=7pi/6-6pi/6 = (7pi-6pi) / 6=pi/6

Then

sin (-7pi/6) = sin (pi/6) →sin (-7pi/6) = 1/2

cos (-7pi/6) = - cos (pi/6) →cos (-7pi/6) = - sqrt (3) / 2

csc (-7pi/6) = 1/sin (-7pi/6) = 1 / (1/2) = 1 (2/1) = 2/1→csc (-7pi/6) = 2

sec (-7pi/6) = 1/cos (-7pi/6) = 1 / (-sqrt (3) / 2) = - 1 (2/sqrt (3)) = - 2/sqrt (3) →

sec (-7pi/6) = - [2/sqrt (3) ]*sqrt (3) / sqrt (3) = - 2sqrt (3) / [sqrt (3) ]^2→

sec (-7pi/6) = - 2sqrt (3) / 3

tan (-7pi/6) = sin (-7pi/6) / cos (-7pi/6) = (1/2) / (-sqrt (3) / 2) = - (1/2) * (2/sqrt (3)) →

tan (-7pi/6) = - 2/[2sqrt (3) ]=-1/sqrt (3) = - [1/sqrt (3) ]*[sqrt (3) / sqrt (3) ]→

tan (-7pi/6) = - sqrt (3) / [sqrt (3) ]^2→tan (-7pi/6) = - sqrt (3) / 3

cot (-7pi/6) = cos (-7pi/6) / sin (-7pi/6) = [-sqrt (3) / 2] / (1/2) = - sqrt (3) / 2 * (2/1) →

cot (-7pi/6) = - 2sqrt (3) / 2→cot (-7pi/6) = - sqrt (3)

Answers:

sin (-7pi/6) = 1/2

cos (-7pi/6) = - sqrt (3) / 2

tan (-7pi/6) = - sqrt (3) / 3

csc (-7pi/6) = 2

sec (-7pi/6) = - 2*sqrt (3) / 2

cot (-7pi/6) = - sqrt (3)