Prove that
cosecθcosecθ−sinθ≡sec2θ\displaystyle \frac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta - \sin\theta} \equiv \sec^2\thetacosecθ−sinθcosecθ≡sec2θ
Hence solve, for 0<θ<2π0 < \theta < 2\pi0<θ<2π, the equation
cosecθcosecθ−sinθ=2tanθ\displaystyle \frac{\operatorname{cosec}\theta}{\operatorname{cosec}\theta - \sin\theta} = 2\tan\thetacosecθ−sinθcosecθ=2tanθ
Give your answers in terms of π\piπ.
225 exam-style questions on WJEC A Level Maths 3.5 Trigonometry (A-level only), covering 3.5.1 Trigonometry (A-level only), 3.5.2 Trigonometry (A-level only), 3.5.3 Trigonometry (A-level only), 3.5.4 Trigonometry (A-level only), 3.5.5 Trigonometry (A-level only), 3.5.6 Trigonometry (A-level only), 3.5.7 Trigonometry (A-level only), and 3.5.8 Trigonometry (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.