In the absence of any specification for the range, like# in [0, 2pi]#, the inverse trigonometric functions are many-valued. General solution has to be given as the answer.
Here cotangent is #-sqrt 3<0 to# sine and cosine of the angle will be of opposite signs,
So, the principal value could be chosen as either the angle in the 2nd quadrant or in the fourth quadrant, in which sine and cosine have opposite signs.
#cot(pi-pi/6) = cot( 2 pi-pi/6 )=-cot pi/6 = -sqrt 3#.
So the principal value of #arc cot (-sqrt 3)# could be either #(5pi)/6 or (11pi)/6#.
The general value is #npi + (5pi)/6, n = 0. +-1, +-2, +-3...#, representing the same two opposite directions #theta=(5pi)/6, (11pi)/6#, in a cycle...
