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If sinθcos3θ+sin3θcos9θ+sin9θcos27θ+sin27θcos81θ=AtanBθ−tanCθ  then B−CA=                 A>0,θ≠2n+1π2,n∈Z

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answer is 164.

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Detailed Solution

We havesinθcos3θ=12sin2θcosθcos3θ=12tan3θ−tanθSimilarly sin3θcos9θ=12tan9θ−tan3θsin9θcos27θ=12tan27θ−tan9θ and sin27θcos81θ=12tan81θ−tan27θ∴sinθcos3θ+sin3θcos9θ+sin9θcos27θ+sin27θcos81θ=12tan81θ−tanθ⇒B−CA=81+11/2=164
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