If  sinθcos3θ+sin3θcos9θ+sin9θcos27θ=AtanBθ−tanCθ, then B+CA=                             A>0,θ≠2n+1π2,n∈Z

We  have  sinθcos3θ=12sin2θcosθcos3θ 12sin(3θ-θ)cosθcos3θ=12sin3θcosθ-cos3θsinθcosθcos3θ  sinθcos3θ=12tan3θ−tanθ   -----(1)Similarly   sin3θcos9θ=12tan9θ−tan3θ ----(2)and           sin9θcos27θ=12tan27θ−tan9θ  ----(3)adding (1) ,(2) ,(3)∴   sinθcos3θ+sin3θcos9θ+sin9θcos27θ=12tan27θ−tanθ=AtanBθ−tanCθ∴B+CA=27+112=56