Additional Mathematics 0606 / Trigonometry / 2.162.16: Simplify tan over sec and solve0606/22/M/J/19 — Question 9 · 4 marksMark as done · Save for later · Show all solutions(b)(i)Show that, for −π2<y<π2-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}−2π<y<2π, 4tany1+tan2y\dfrac{4\tan y}{\sqrt{1 + \tan^{2} y}}1+tan2y4tany can be written in the form asinya\sin yasiny, where aaa is an integer.[3]▸ Answer(b)(ii)Hence solve 4tany1+tan2y+3=0\dfrac{4\tan y}{\sqrt{1 + \tan^{2} y}} + 3 = 01+tan2y4tany+3=0 for −π2<y<π2-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}−2π<y<2π radians.[1]▸ Answer▸ Official mark scheme← 2.5: Cosecant–cotangent identity and tan equation2.25: Sketch, period and amplitude of sine graph →
