Expanding the LHS and simplifying the RHS, we have $5$tan$^2A+4$tan$^2B+20$tan$^2A$tan$^2B=8\sqrt5$tan$A$tan$B$. Subtracting $8\sqrt5$tan$A$tan$B$ from both sides, we obtain $5$tan$^2A-4\sqrt5$tan$A$tan$B+4$tan$^2B+20+$tan$^2A$tan$^2B-4$tan$A$tan$B=(2$tan$B-\sqrt5$tan$A)^2+($tan$A$tan$B-2\sqrt5)^2=0$, which is possible iff tan$A$tan$B=2\sqrt5$ and $2$tan$B=\sqrt5$tan$A$. Solving for tan$A$ and tan$B$, we obtain tan$A=2$ and tan$B=\sqrt5$, as tan$A$ and tan$B$ must both be positive since $0<A,B<\frac{\pi}{2}$. Thus, we have sin$A=2$cos$A$ and sin$B=\sqrt5$cos$B$, and since sin$^2A+$cos$^2A=1$ and, similarly, sin$^2B+$cos$^2B=1$, we can solve for cos$A$ and sin$B$, which yield $\frac{\sqrt5}{5}$ and $\frac{\sqrt {30}}{6}$, respectively, thus the only possible value of the product cos$A$sin$B$ is ${\frac{\sqrt5}{5}}\cdot{\frac{\sqrt {30}}{6}}=\frac{\sqrt 6}{6}$ $\blacksquare$

Expanding the LHS and simplifying the RHS, we have $5$tan$^2A+4$tan$^2B+20$tan$^2A$tan$^2B=8\sqrt5$tan$A$tan$B$. Subtracting $8\sqrt5$tan$A$tan$B$ from both sides, we obtain $5$tan$^2A-4\sqrt5$tan$A$tan$B+4$tan$^2B+20+$tan$^2A$tan$^2B-4$tan$A$tan$B=(2$tan$B-\sqrt5$tan$A)^2+($tan$A$tan$B-2\sqrt5)^2=0$, which is possible iff tan$A$tan$B=2\sqrt5$ and $2$tan$B=\sqrt5$tan$A$. Solving for tan$A$ and tan$B$, we obtain tan$A=2$ and tan$B=\sqrt5$, as tan$A$ and tan$B$ must both be positive since $0<A,B<\frac{\pi}{2}$. Thus, we have sin$A=2$cos$A$ and sin$B=\sqrt5$cos$B$, and since sin$^2A+$cos$^2A=1$ and, similarly, sin$^2B+$cos$^2B=1$, we can solve for cos$A$ and sin$B$, which yield $\frac{\sqrt5}{5}$ and $\frac{\sqrt 30}{6}$, respectively, thus the only possible value of the product cos$A$sin$B$ is ${\frac{\sqrt5}{5}}\cdot{\frac{\sqrt 30}{6}}=\frac{\sqrt 6}{6}$ $\blacksquare$

By AMGM, we have $4+\tan^2(A) \ge 2\sqrt{4\tan^2(A)}$ and $5+\tan^2(B) \ge 2\sqrt{5\tan^2(B)}$ Multiplying the 2 equations gives $(4 + \tan^2(A))(5 + \tan^2(B)) \ge \sqrt{320}\tan(A)\tan(B)$ From the problem statement, equality holds so $4 = \tan^2(A)$ and $5 = \tan^2(B)$, then proceed as above