Prove that 11+cosθ+11−cosθ≡2sin2θ\dfrac{1}{1 + \cos\theta} + \dfrac{1}{1 - \cos\theta} \equiv \dfrac{2}{\sin^2\theta}1+cosθ1+1−cosθ1≡sin2θ2 where sinθ≠0\sin\theta \neq 0sinθ=0.