If a·co$s^2$θ + b·si$n^2$θ = $c$, then ta$n^2$θ equals:

Divide the whole equation by co$s^2$θ: $a+b$·ta$n^2$θ = $c$·se$c^2$θ = $c$(1 + ta$n^2$θ). Collecting the ta$n^2$θ terms gives ta$n^2$θ$(b-c)=c-a$, so ta$n^2$θ = $(c-a)/(b-c)=(a-c)/(c-b)$ — the two middle choices each flip exactly one sign. Sanity check with $a=3,b=1$, θ = 45°: then $c=2$ and (3 − 2)/(2 − 1) = 1 = ta$n^2$45°.