Additional Mathematics 0606 / Trigonometry / 2.52.5: Cosecant–cotangent identity and tan equation0606/21/M/J/19 — Question 11 · 9 marksMark as done · Save for later · Show all solutions(a)(i)Show that cscθ−cotθsinθ=11+cosθ\dfrac{\csc \theta - \cot \theta}{\sin \theta} = \dfrac{1}{1 + \cos \theta}sinθcscθ−cotθ=1+cosθ1.[4]▸ Answer(a)(ii)Hence solve cscθ−cotθsinθ=52\dfrac{\csc \theta - \cot \theta}{\sin \theta} = \tfrac{5}{2}sinθcscθ−cotθ=25 for 180°<θ<360°180^{\degree} < \theta < 360^{\degree}180°<θ<360°.[2]▸ Answer(b)Solve tan(3ϕ−4)=−12\tan(3\phi - 4) = -\tfrac{1}{2}tan(3ϕ−4)=−21 for 0≤ϕ≤π20 \le \phi \le \tfrac{\pi}{2}0≤ϕ≤2π radians.[3]▸ Answer▸ Official mark scheme← 1.27: Secant identity and squared sine equation2.16: Simplify tan over sec and solve →
