cos2θ+sin2θ=1; 1+tan2θ=sec2θ; cot2θ+1=cosec2θ (trig identities).
So the expression becomes:
1+2cot2θ+1+1+2tan2θ=3+2(tan2θ+cot2θ).
The expression is minimum when tan2θ+cot2θ is minimum.
cotθ=1/tanθ=cosθ/sinθ,
tan2θ+cot2θ=(tanθ+cotθ)2-2tanθcotθ=
[(sin2θ+cos2θ)/(sinθcosθ)]2-2=1/(sinθcosθ)2-2=4/sin2(2θ)-2.
The minimum value of 4/sin2(2θ)-2 is when sin(2θ)=1, so 4/sin2(2θ)-2=2.
The minimum value of tan2θ+cot2θ is therefore 2, making the minimum value of the expression 3+2×2=7.