Let's first figure out where the angle #(7pi)/6# puts us. With #(6pi)/6=pi# and so half way around a circle (the arm sweeps through Q1 and Q2) and then moves #pi/6# further. And so we're in Q3 with a reference angle of #pi/6#.
What do we know about a triangle with a reference angle of #pi/6#? It's a special triangle, known in degrees as a 30-60-90 triangle and has ratios of the small, medium, and long sides as 1, #sqrt3#, 2. We know it's the short side that is the opposite side, the medium that is the adjacent, and the hypotenuse that is the long side. We also know that the adjacent and opposite sides are going to have negative values.
So the ratios work out to be:
#sin="opp"/"hyp"=-1/2#
#cos="adj"/"hyp"=-sqrt3/2#
#tan="opp"/"adj"=(-1)/-sqrt3=sqrt3/3#
#csc="hyp"/"opp"=-2/1=-2#
#sec="hyp"/"adj"=-2/sqrt3=-(2sqrt3)/3#
#cot="adj"/"opp"=(-sqrt3)/-1=sqrt3#
