(1+sinθ)(1−sinθ)=1−sinθ+sinθ−sin2θ(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin \theta + \sin \theta - \sin^{2} \theta(1+sinθ)(1−sinθ)=1−sinθ+sinθ−sin2θ
=1−sin2θ= 1 - \sin^{2} \theta=1−sin2θ
Recall, cos2θ+sin2θ=1\cos^{2} \theta + \sin^{2} \theta = 1cos2θ+sin2θ=1
∴1−sin2θ=cos2θ\therefore 1 - \sin^{2} \theta = \cos^{2} \theta∴1−sin2θ=cos2θ.
