For an acute angle θ, tanθ = $2ab/(a^2-b^2)$ where $a>b>0$. Then sinθ equals:

Read the tangent as a right triangle with opposite side $2ab$ and adjacent side $a^2-b^2$. The hypotenuse is √$(4a^2{b}^2+(a^2-b^2{)}^2)$ = √$((a^2+b^2{)}^2)=a^2+b^2$, so sinθ = $2ab/(a^2+b^2)$. With $a=2$ and $b=1$ this says tanθ = 4/3 and sinθ = 4/5, the familiar 3-4-5 triangle. The choice $(a^2-b^2)/(a^2+b^2)$ is cosθ, $(a^2+b^2)/(2ab)$ is cosecθ, and $(a^2-b^2)/(2ab)$ is cotθ.