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Question

Show that sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ =2sec2θ.

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Solution

sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ

=(sinθ+cosθ)2+(sinθcosθ)2sin2θcos2θ

=sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθsin2θcos2θ

=1+1sin2θcos2θ

=2(cos2θsin2θ)

=2cos2θ

=2sec2θ

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