If $x$ = tanθ + sinθ and $y$ = tanθ − sinθ, which relation holds?

$x^2-y^2=4$·tanθ·sinθ, while $xy$ = ta$n^2$θ − si$n^2$θ = si$n^2$θ(se$c^2$θ − 1) = si$n^2$θ·ta$n^2$θ, so √$(xy)=$ sinθ·tanθ. Hence $x^2-y^2=4$√$(xy)$, and squaring gives $(x^2-y^2{)}^2=16xy$.