Additional Mathematics 0606 / Trigonometry / 2.192.19: Reciprocal sine identity and equation0606/22/O/N/18 — Question 8 · 8 marksMark as done · Save for later · Show all solutions(i)Show that 11−sinx−11+sinx=2tanxsecx\dfrac{1}{1 - \sin x} - \dfrac{1}{1 + \sin x} = 2\tan x \sec x1−sinx1−1+sinx1=2tanxsecx.[4]▸ Answer(ii)Hence solve the equation 11−sinx−11+sinx=cscx\dfrac{1}{1 - \sin x} - \dfrac{1}{1 + \sin x} = \csc x1−sinx1−1+sinx1=cscx for 0°≤x≤360°0^{\degree} \le x \le 360^{\degree}0°≤x≤360°.[4]▸ Answer▸ Official mark scheme← 2.9: Cosecant–cotangent identity and equation2.30: Sine, sec–cosec and cot–tan equations →
