I think this is meant to read: prove sec2θ+cosec2θ≥4.
1/cos2θ+1/sin2θ=(sin2θ+cos2θ)/(sinθcosθ)2=
1/(sinθcosθ)2.
sin(2θ)=2sinθcosθ, so sinθcosθ=½sin(2θ) and (sinθcosθ)2=½2sin2(2θ)=¼sin2(2θ).
The highest magnitude |sin(2θ)|=1 when 2θ=90° or 270°, for example. So maximum sin2(2θ)=1, so maximum (sinθcosθ)2=¼. The reciprocal of this therefore has a minimum value of 1/¼=4, therefore:
1/(sinθcosθ)2≥4, and sec2θ+cosec2θ≥4 QED