Zadanie 11b, Paper 1, maj 2026, TZ1, Matematyka AA IB HL

Using (a), multiply $\tfrac12(\cot\theta-1/\cot\theta)=\tfrac16\cot\theta(\cot^2\theta-1)$ by 6cotθ. This gives $3(\cot^2\theta-1)=\cot^2\theta(\cot^2\theta-1)$, so $\cot^4\theta-4\cot^2\theta+3=0$. Factor as $(\cot^2\theta-1)(\cot^2\theta-3)=0$. Thus cotθ=±1 or ±√3, giving θ = π/6, π/4, 3π/4, 5π/6. All satisfy the original equation and domain.

M1: use (a) and clear denominators to form a quadratic in cot²θ or tan²θ. M1: solve that three-term quadratic. A2: all four correct cotangent/tangent values; award A1 for two or three. A2: all four exact angles; award A1 for two or three. Ignore additional answers for these source-compatible answer marks. If cot²θ−1 is cancelled without retaining its zero cases, cap at 3/6: first M1, second M0, at most one intermediate A1 and one final A1.