Additional Mathematics 0606 / Trigonometry / 2.202.20: Tan–sec sum identity and sine equation0606/22/O/N/19 — Question 6 · 9 marksMark as done · Save for later · Show all solutions(i)Show that tanx1+secx+1+secxtanx=2sinx\dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} = \dfrac{2}{\sin x}1+secxtanx+tanx1+secx=sinx2.[5]▸ Answer(ii)Hence solve the equation tanx1+secx+1+secxtanx=1+3sinx\dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} = 1 + 3\sin x1+secxtanx+tanx1+secx=1+3sinx for 0°≤x≤180°0^{\degree} \le x \le 180^{\degree}0°≤x≤180°.[4]▸ Answer▸ Official mark scheme← 2.11: Cubic in sin from sec–tan equation2.31: Cosecant–cotangent over 1 − cos →