If $p=a$·secθ + b·tanθ and $q=a$·tanθ + b·secθ, then $p^2-q^2$ equals:

Expanding, $p^2=a^2$se$c^2$θ + $2ab$·secθ·tanθ + $b^2$ta$n^2$θ and $q^2=a^2$ta$n^2$θ + $2ab$·secθ·tanθ + $b^2$se$c^2$θ. The middle terms are identical and cancel on subtraction, leaving $a^2$(se$c^2$θ − ta$n^2$θ) − $b^2$(se$c^2$θ − ta$n^2$θ) = $(a^2-b^2)\times 1=a^2-b^2$. Check with $a=5,b=3$, θ = 64°: $p=11.404+6.151=17.555$ and $q=10.252+6.842=17.094$, and 308.2 − 292.2 = 16.0 = 25 − 9. Using se$c^2$θ + ta$n^2$θ instead of the difference gives $a^2+b^2$; expanding $p^2-q^2$ as $(p-q{)}^2$ gives $(a-b{)}^2$; and 1 forgets the a, $b$ factors entirely.