Prove the identity
1−cos2xsin2x≡tanx\displaystyle \frac{1 - \cos 2x}{\sin 2x} \equiv \tan xsin2x1−cos2x≡tanx
Hence solve, for 0≤θ<2π0 \leq \theta < 2\pi0≤θ<2π, the equation
1−cos2θ=sin2θ1 - \cos 2\theta = \sin 2\theta1−cos2θ=sin2θ
Give your answers in terms of π\piπ.