Reciprocal double-angle identity used to solve an equation, differentiated, and applied to an isosceles triangle

Prove that \(\operatorname{cosec}2\theta-\cot2\theta=\tan\theta\), where \(\sin2\theta\ne0\).[3]

Hence, solve the equation \(3(\operatorname{cosec}2\theta-\cot2\theta)^2+3\cot^2\theta=10\) for \(0<\theta<\pi\), \(\theta\ne\frac\pi2\).[6]

Differentiate the identity proved in part (a) with respect to \(\theta\), and hence show that \(2\operatorname{cosec}2\theta\tan\theta=\sec^2\theta\).[5]

The diagram below shows the isosceles triangle \(DEF\), in which \(DE=DF=6\sec\theta\) and \(E\hat{D}F=2\theta\), where \(0<\theta<\frac\pi2\). All lengths are in centimetres.

The length of \(EF\) is \(\ell\text{ cm}\) and the area of triangle \(DEF\) is \(T\text{ cm}^2\).

Find the ratio \(T:\ell\), giving your answer in the form \(n:1\), where \(n\in\mathbb{Z}^+\).[5]