(sinΘ+cosΘ)2==sin2Θ+2sinΘ∗cosΘ+cos2Θ==sin2Θ+cos2Θ+sin2Θ==1+sin2Θ(sin\varTheta +cos\varTheta)^2 =\\ = sin^2\varTheta +2sin\varTheta*cos\varTheta +cos^2\varTheta=\\ = sin^2\varTheta+cos^2\varTheta +sin2\varTheta = \\ =1 + sin2\varTheta(sinΘ+cosΘ)2==sin2Θ+2sinΘ∗cosΘ+cos2Θ==sin2Θ+cos2Θ+sin2Θ==1+sin2Θ
answer: 1+sin2Θ1 + sin2\varTheta1+sin2Θ
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