Additional Mathematics 0606 / Trigonometry / 3.943.94: Tangent–sine identity0606/23/O/N/23 — Question 8 · 10 marksMark as done · Save for later · Show all solutions(a)Show that sin xtan x−1−cos xtan x+1=cos xsin2x−cos2x\dfrac{\mathrm{sin}\,x}{\mathrm{tan}\,x-1}-\dfrac{\mathrm{cos}\,x}{\mathrm{tan}\,x+1}=\dfrac{\mathrm{cos}\,x}{\mathrm{sin}^{2}x-\mathrm{cos}^{2}x}tanx−1sinx−tanx+1cosx=sin2x−cos2xcosx.[5]▸ Answer(b)Hence solve the equation sin xtan x−1−cos xtan x+1=1\dfrac{\mathrm{sin}\,x}{\mathrm{tan}\,x-1}-\dfrac{\mathrm{cos}\,x}{\mathrm{tan}\,x+1}=1tanx−1sinx−tanx+1cosx=1 for 0∘⩽x⩽360∘0^{\circ}\leqslant x\leqslant 360^{\circ}0∘⩽x⩽360∘.[5]▸ Answer▸ Official mark scheme← 3.93: Triangle with given angles (no calculator)3.29: Triangles with exact trig ratios →
