The value of $$\frac{\tan13^{\circ}\tan36^{\circ}\tan45^{\circ}\tan54^{\circ}\tan77^{\circ}}{2\sec^260^{\circ}(\sin^260^{\circ}-3\cos60^{\circ}+2)}$$ is:

$$\frac{\tan13^{\circ}\tan36^{\circ}\tan45^{\circ}\tan54^{\circ}\tan77^{\circ}}{2\sec^260^{\circ}(\sin^260^{\circ}-3\cos60^{\circ}+2)}=\frac{\tan13^{\circ}\tan36^{\circ}\left(1\right)\tan\left(90-36^{\circ}\right)\tan\left(90-13^{\circ\ }\right)}{2\left(2\right)^2\left(\left(\frac{\sqrt{3}}{2}\right)^2-3\left(\frac{1}{2}\right)+2\right)}$$

$$=\frac{\tan13^{\circ}\tan36^{\circ}\cot36^{\circ}\cot13^{\circ\ }}{2\left(4\right)\left(\frac{3}{4}-\frac{3}{2}+2\right)}$$

$$=\frac{1}{8\left(\frac{3-6+8}{4}\right)}$$

$$=\frac{1}{2\left(5\right)}$$

$$=\frac{1}{10}$$

Hence, the correct answer is Option A