Question

prove that cot (180-θ) = -cot θ

Answer 1

Use cot(A-B) expansion

Answer 2

it is rule that cot (180-θ) = -cot θ because cot theta is negative in second quadrant.

Answer 3

(180-θ) need not be in the 2nd quadrant, though....?? This is a general proof of an identity. Like @AravindG said, use cot(A-B). If you haven't covered that as an identity, just use the fact that
\(\Large \cot x=\dfrac{\cos x}{\sin x}\) and so \(\Large \cot (180-\theta)=\dfrac{\cos (180-\theta)}{\sin (180-\theta)}\)

Then use the sum and difference identities to expand the num'r and den'r, simplify, and everything falls into place. :)