For every acute angle θ, the value of co$t^2$θ − co$s^2$θ − co$t^2$θ·co$s^2$θ is:

Take co$t^2$θ out of the first and third terms: co$t^2$θ(1 − co$s^2$θ) − co$s^2$θ = co$t^2$θ·si$n^2$θ − co$s^2$θ, and since cotθ·sinθ = cosθ the first product is exactly co$s^2$θ, leaving co$s^2$θ − co$s^2$θ = 0 for every θ. (Grouping the other way is just as quick: co$t^2$θ − co$s^2$θ(1 + co$t^2$θ) = co$t^2$θ − co$s^2$θ·cose$c^2$θ = co$t^2$θ − co$t^2$θ = 0.) Check at θ = 40°: 1.42027 − 0.58682 − 0.83345 = 0.0000. Simplifying co$t^2$θ·si$n^2$θ to co$s^2$θ and then forgetting to subtract the remaining −co$s^2$θ gives co$s^2$θ; ignoring the last two terms altogether gives co$t^2$θ; and 1 comes from expecting every tidy trigonometric expression to collapse to the Pythagorean 1.