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Question

If sin3θcos3θsinθcosθcosθ1+cot2θ2tanθcotθ=1 where θ [0,2π] then

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Solution

sin3θcos3θsinθcosθ+cosθ1+cos2θ2tanθcotθ=1(sinθcosθ)(sin2θ+cos2θ+sinθcosθ)(sinθcosθ)+cosθcosecθ2=11+sinθ.cosθ+sinθcosθ=12sinθ.cosθ=0sin2θ=02θ=nπθ=nπ2θ=π2orπor3π2

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