For a given angle θ, the two numbers secθ + tanθ and secθ − tanθ are the roots of the quadratic $x^2-px+q=0$. Then $p^2-4q$ equals:

The sum of the roots is $p=2$secθ and their product is $q$ = se$c^2$θ − ta$n^2$θ = 1. So $p^2-4q=4$se$c^2$θ − 4 = 4(se$c^2$θ − 1) = 4ta$n^2$θ. Check at θ = 60°: the roots are 3.7321 and 0.2679, so $p=4$ and $q=1$, and $16-4=12=4\times 3=4$ta$n^2$60°. Forgetting the $-4q$ term leaves 4se$c^2$θ; a discriminant of 0 would mean equal roots, which happens only at θ = 0°; and 4 would require $q=0$, i.e. one root equal to zero, which secθ ± tanθ never is.