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Question

sinθ+cosθsinθ-cosθ+sinθ-cosθsinθ+cosθ=2(sin2θ-cos2θ)=2(2sin2θ-1)

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Solution

Wehavesinθ+cosθsinθ−cosθ+sinθ−cosθsinθ+cosθ=(sinθ+cosθ)2+(sinθ−cosθ)2(sinθ−cosθ)(sinθ+cosθ)=sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ−2sinθcosθsin2θ−cos2θ=1+1sin2θ−cos2θ=2sin2θ−cos2θAgain,2sin2θ−cos2θ=2sin2θ−(1−sin2θ)=22sin2θ−1

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