VMO

Note that $x\ne \frac{k\pi }{2}$, $y\ne \frac{m\pi }{2}$ ($k,m\in \mathbb{Z}$) and $xy>0$. From the given condition, one can get $$A=20\sqrt{\frac{xy}{(x+y{)}^2}},$$ where $A$ equals to the product of two left hand sides of the given system. Using the Cauchy-Schwarz and AM-GM inequality $$\begin{array}{rl}& ({\sin }^{2}x+\frac{1}{{\sin }^{2}x})({\cos }^{2}x+\frac{1}{{\cos }^{2}x})\\ & \ge (|\sin x\cos x|+\frac{1}{|\sin x\cos x|}{)}^2\\ & =(\frac{|\sin 2x|}{2}+\frac{1}{2|\sin 2x|}+\frac{3}{2|\sin 2x|}{)}^2\\ & \ge (1+\frac{3}{2}{)}^2=(\frac{5}{2}{)}^2.\end{array}$$ Similarly, we have $$({\sin }^{2}y+\frac{1}{{\sin }^{2}y})({\cos }^{2}y+\frac{1}{{\cos }^{2}y})\ge (\frac{5}{2}{)}^2$$ Therefore, applying AM-GM inequality, we obtain $$A\ge 4\sqrt[4]{{\left(\frac{5}{2}\right)}^4}=10\ge 20\sqrt{\frac{xy}{(x+y{)}^2}}$$ The equality occurs if and only if $|\sin 2x|=1$ and $x=y=\frac{\pi }{4}+\frac{k\pi }{2}$ where $k\in \mathbb{Z}$. It is easy to check that these solutions satisfy the given system. Thus, $x=y=\frac{\pi }{4}+\frac{k\pi }{2}$ where $k\in \mathbb{Z}$ are all solutions of the given system. $\square$