Additional Mathematics 0606 / Trigonometry / 2.172.17: Cotangent quadratic and fourth-power identity0606/22/M/J/20 — Question 8 · 9 marksMark as done · Save for later · Show all solutions(a)Solve 3cot2x−14cscx−2=03\cot^{2} x - 14\csc x - 2 = 03cot2x−14cscx−2=0 for 0°<x<360°0^{\degree} < x < 360^{\degree}0°<x<360°.[5]▸ Answer(b)Show that sin4y−cos4ycoty=tany−2cosysiny\dfrac{\sin^{4} y - \cos^{4} y}{\cot y} = \tan y - 2\cos y \sin ycotysin4y−cos4y=tany−2cosysiny.[4]▸ Answer▸ Official mark scheme← 1.15: Tangent equation and graph sketch2.26: Sine curve constants and tangent sketch →
