For sinθ ≠ ±cosθ, the expression (si$n^6$θ − co$s^6$θ)/(si$n^2$θ − co$s^2$θ) simplifies to:

Put $a$ = si$n^2$θ and $b$ = co$s^2$θ, so $a+b=1$. The quotient $(a^3-b^3)/(a-b)$ is $a^2+ab+b^2=(a+b{)}^2-ab=1-$ si$n^2$θ·co$s^2$θ. Check at θ = 30°: $(1/64-27/64)/(1/4-3/4)=0.8125$ and $1-(1/4)(3/4)=0.8125$. Using the sum-of-cubes bracket $a^2-ab+b^2$ gives 1 + si$n^2$θ·co$s^2$θ; the remaining two are the standard values of si$n^4$θ + co$s^4$θ and si$n^6$θ + co$s^6$θ, neither of which is what this quotient equals.