Given that sinθ≠±1\sin \theta \neq \pm 1sinθ=±1, prove the identity 11−sinθ+11+sinθ≡2sec2θ\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} \equiv 2 \sec^2 \theta1−sinθ1+1+sinθ1≡2sec2θ
Hence, find the set of values of kkk for which the equation 11−sinθ+11+sinθ=k\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = k1−sinθ1+1+sinθ1=k has real solutions. Fully justify your answer.
Given that θ\thetaθ is in the second quadrant (reflex angle between 90∘90^\circ90∘ and 180∘180^\circ180∘) and 11−sinθ+11+sinθ=10\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = 101−sinθ1+1+sinθ1=10 find the exact value of tanθ\tan \thetatanθ.