IB AA HL May 2026 P1 TZA Q11

Prove the identity \(\cot2\theta=\frac12\left(\cot\theta-\frac{1}{\cot\theta}\right)\).[3]

Hence, solve the equation \(\cot2\theta=\frac32\cot\theta(\cot^2\theta-1)\) for \(0<\theta<\pi\) where \(\theta\ne\frac\pi2\).[6]

By differentiating both sides of the identity in part (a), show that \(4\operatorname{cosec}^2 2\theta=\operatorname{cosec}^2\theta+\sec^2\theta\).[5]

Consider the right-angled triangle \(\triangle PQR\), where \(R\hat{P}Q=\frac\pi2\).

\(PR=\operatorname{cosec}\theta\), \(QR=2\operatorname{cosec}2\theta\), where \(0<\theta<\frac\pi2\) and all lengths are given in metres.

\(QR\) has length \(L\) metres and \(\triangle PQR\) has area \(A\) square metres.

Calculate the ratio \(L:A\) in the form \(n:1\), where \(n\in\mathbb{Z}^+\).[5]