Additional Mathematics 0606 / Trigonometry / 1.271.27: Secant identity and squared sine equation0606/13/M/J/19 — Question 6 · 10 marksMark as done · Save for later · Show all solutions(a)(i)Show that secθ−tanθcscθ=cosθ\sec \theta - \tfrac{\tan \theta}{\csc \theta} = \cos \thetasecθ−cscθtanθ=cosθ.[3]▸ Answer(a)(ii)Solve sec2θ−tan2θcsc2θ=32\sec 2\theta - \tfrac{\tan 2\theta}{\csc 2\theta} = \tfrac{\sqrt{3}}{2}sec2θ−csc2θtan2θ=23 for 0∘≤θ≤180∘0^{\circ} \le \theta \le 180^{\circ}0∘≤θ≤180∘.[3]▸ Answer(b)Solve 2sin2(ϕ+π3)=12\sin^{2}\left(\phi + \tfrac{\pi}{3}\right) = 12sin2(ϕ+3π)=1 for 0<ϕ<2π0 < \phi < 2\pi0<ϕ<2π radians.[4]▸ Answer▸ Official mark scheme← 1.4: Amplitude, period and sketch of a sine graph2.5: Cosecant–cotangent identity and tan equation →
