Prove that
tanϕ+cotϕ≡2csc2ϕ \tan \phi + \cot \phi \equiv 2 \csc 2\phi tanϕ+cotϕ≡2csc2ϕfor ϕ≠nπ2,n∈Z\phi \neq \frac{n\pi}{2}, n \in \mathbb{Z}ϕ=2nπ,n∈Z.
Using the identity in part (a), or otherwise, prove that
cot2ϕ−tan2ϕ≡4cot2ϕcsc2ϕ \cot^2 \phi - \tan^2 \phi \equiv 4 \cot 2\phi \csc 2\phi cot2ϕ−tan2ϕ≡4cot2ϕcsc2ϕfor ϕ≠nπ2,n∈Z\phi \neq \frac{n\pi}{2}, n \in \mathbb{Z}ϕ=2nπ,n∈Z.
Hence solve, for −π2<α<π2-\frac{\pi}{2} < \alpha < \frac{\pi}{2}−2π<α<2π,
4cot2αcsc2α=3tan2α+1 4 \cot 2\alpha \csc 2\alpha = 3 \tan^2 \alpha + 1 4cot2αcsc2α=3tan2α+1giving your answers to 2 decimal places.