For every θ at which it is defined, $(cosec\theta -cot\theta {)}^2$ is equal to:

cosecθ − cotθ = (1 − cosθ)/sinθ, so the square is $(1-cos\theta {)}^2$/si$n^2$θ. Writing si$n^2$θ as (1 − cosθ)(1 + cosθ) and cancelling one factor leaves (1 − cosθ)/(1 + cosθ). Check at θ = 60°: $(1.1547-0.5774{)}^2=0.3333$ and $(1-0.5)/(1+0.5)=0.3333$. The identity cose$c^2$θ − $co$$t^2$θ = 1 concerns the difference of the squares, not the square of the difference, so the answer is not 1; the sine version belongs to $(sec\theta -tan\theta {)}^2$.