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Question

1cos2θ+sin2θ1+cos2θ+sin2θ=tanθ

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Solution

1cos2θ+sin2θ1+cos2θ+sin2θ=tanθ=1cos2θ+sin2θ+2sinθcosθ1cos2θsin2θ+2sinθcosθ
=(1cos2θ)+sin2θ+2sinθcosθ(1sin2θ)+cos2θ+2sinθcosθ
=2sin2θ+2sinθcosθ2cos2θ+2sinθcosθ=2sinθ2cosθ(sinθ+cosθcosθ+sinθ)=tanθ

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