If we take a right triangle ABC where B is the right angle, and the sides are a, b, c such that tanθ=a/c, cosθ=c/√(a2+c2)=c/b, sinθ=a/√(a2+c2), where b=√(a2+c2) (Pythagoras).
cotθ=c/a, cot2θ=c2/a2; cos2θ=c2/(a2+c2)=c2/b2.
So we have:
cot2θ-cos2θ=c2/a2-c2/(a2+c2)=c2(1/a2-1/(a2+c2))=c2(a2+c2-a2)/(a2(a2+c2))=c4/(ab)2.
c4/(ab)2=(c/a)2(c/b)2=cot2θcos2θ.
Therefore: cot2θ-cos2θ=cot2θcos2θ QED