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Question

Solve
sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ=2sec2θtan2θ1

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Solution

sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ=2sec2θtan2θ1
L.H.S
=(sinθ+cosθ)2+(sinθcosθ)2(sinθ+cosθ)(sinθcosθ)
=sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθsin2θcos2θ
=sin2θ+cos2θ+sin2θ+cos2θsin2θcos2θ
=2sin2θ+2cos2θsin2θcos2θ
=2(sin2θ+cos2θ)(sin2θcos2θ)
Now, RHS
=2sec2θtan2θ1=21cos2θsin2θcos2θ1
=2/cos2θsin2θcos2θ/cos2θ
=2sin2θcos2θ
L.H.S=R.H.S

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