If 1−tan⁡θtan⁡θ11tan⁡θ−tan⁡θ1−1=a−bba then

a−bba=1−tan⁡θtan⁡θ1cos2⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θcos2⁡θ                =cos⁡2θ−sin⁡2θsin⁡2θcos⁡2θ               ⇒a=cos⁡2θ,b=sin⁡2θ