Prove that sec2θ≡cscθcscθ−sinθ\displaystyle \sec^2 \theta \equiv \frac{\csc \theta}{\csc \theta - \sin \theta}sec2θ≡cscθ−sinθcscθ
Hence verify, for 0<θ<2π0 < \theta < 2\pi0<θ<2π, that θ=π4\displaystyle \theta=\frac{\pi}{4}θ=4π and θ=5π4\displaystyle \theta=\frac{5\pi}{4}θ=45π satisfy cscθcscθ−sinθ=2tanθ\displaystyle \frac{\csc \theta}{\csc \theta - \sin \theta} = 2 \tan \thetacscθ−sinθcscθ=2tanθ
274 exam-style questions on Edexcel A Level Maths Trigonometric Functions, covering 6.1 Secant, Cosecant and Cotangent, 6.2 Graphs of Sec x, Cosec x and Cot x, 6.3 Using Sec x, Cosec x and Cot x, 6.4 Trigonometric Identities, and 6.5 Inverse Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.