cos2θ−1sin2θ\frac{\cos 2\theta - 1}{\sin 2\theta}sin2θcos2θ−1
cos(x+y)=cosxcosy−sinxsiny ⟹ cos2θ=cos2θ−sin2θ\cos (x + y) = \cos x \cos y - \sin x \sin y \implies \cos 2\theta = \cos^{2} \theta - \sin^{2} \thetacos(x+y)=cosxcosy−sinxsiny⟹cos2θ=cos2θ−sin2θ
cos2θ=1−sin2θ ⟹ cos2θ=1−2sin2θ\cos^{2} \theta = 1 - \sin^{2} \theta \implies \cos 2\theta = 1 - 2\sin^{2} \thetacos2θ=1−sin2θ⟹cos2θ=1−2sin2θ
sin2θ=2sinθcosθ\sin 2\theta = 2 \sin \theta \cos \thetasin2θ=2sinθcosθ
∴cos2θ−1sin2θ=1−2sin2θ−12sinθcosθ\therefore \frac{\cos 2\theta - 1}{\sin 2\theta} = \frac{1 - 2\sin^{2}\theta - 1}{2\sin \theta \cos \theta}∴sin2θcos2θ−1=2sinθcosθ1−2sin2θ−1
= −2sin2θ2sinθcosθ=−sinθcosθ\frac{-2\sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{-\sin \theta}{\cos \theta}2sinθcosθ−2sin2θ=cosθ−sinθ
= −tanθ-\tan \theta−tanθ