If tanθ + cotθ = $k$, then se$c^2$θ + cose$c^2$θ written in terms of $k$ is:

se$c^2$θ + cose$c^2$θ = (1 + ta$n^2$θ) + (1 + co$t^2$θ) = 2 + ta$n^2$θ + co$t^2$θ. Squaring the given equation gives ta$n^2$θ + co$t^2$θ = $k^2-2$, since tanθ·cotθ = 1, so the total is $2+(k^2-2)=k^2$. Check at θ = 45°, where $k=2$: se$c^2$45° + cose$c^2$45° = $2+2=4=k^2$. Stopping at ta$n^2$θ + co$t^2$θ leaves $k^2-2$, and adding the 2 instead of cancelling it gives $k^2+2$.