Explanation:

First try it yourself and then see the solution
sin²θ+ cos²θ + sec²θ + cosec²θ + tan²θ + cot²θ
= 1 + sec²θ + cosec²θ + tan²θ + cot²θ…………………….(i)
Now 1 + tan²θ = sec²θ
So (i) = 2sec²θ + cosec²θ + cot²θ…………………………………(ii)
Now cot²θ = cosec²θ – 1
So (ii) = 2sec²θ + cosec²θ + cosec²θ – 1
= 2sec²θ + 2cosec²θ – 1
Now see we had find the min value of 2sec²θ + 2cosec²θ type above
So now min value = 2 + 2 + 2√(2*2) – 1 = 4 + 4 – 1 = 7