Wherever the tangent and the secant are both defined, the difference se$c^4$θ − ta$n^4$θ is always equal to:

Treat it as a difference of two squares: se$c^4$θ − ta$n^4$θ = (se$c^2$θ + ta$n^2$θ)(se$c^2$θ − ta$n^2$θ), and the second bracket is exactly 1, so the whole thing is se$c^2$θ + ta$n^2$θ. Check at θ = 30°: 16/9 − 1/9 = 15/9 = 5/3, and 4/3 + 1/3 = 5/3. Assuming the fourth powers behave like the squares and give 1 is the main trap; substituting se$c^2$θ = 1 + ta$n^2$θ into se$c^2$θ + ta$n^2$θ gives 1 + 2ta$n^2$θ, so dropping that 1 leaves 2ta$n^2$θ = 2/3 here; and se$c^2$θ·ta$n^2$θ = 4/9 multiplies the two brackets that should be added.