smc

Factor each term in the product as a difference of two squares, and group together all the terms that contain a $-$ sign, and all those that contain a $+$ sign. This gives $[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{10})]\cdot [(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4})...(1+\frac{1}{10})]=[\frac{1}{2}\frac{2}{3}\frac{3}{4}...\frac{9}{10}][\frac{3}{2}\frac{4}{3}\frac{5}{4}...\frac{11}{10}]=\frac{1}{10}\cdot \frac{11}{2}=\frac{11}{20}$, which is $\boxed{\frac{11}{20}}$.