$\frac{\cos {20}^{\circ }+8\sin {70}^{\circ }\sin {50}^{\circ }\sin {10}^{\circ }}{{\sin }^2{80}^{\circ }}\text{ is equal to- }$

sin 10° sin 50° sin 70° = cos 20° cos 40° cos 80° = $\frac{\sin \left(2^3\cdot {20}^{\circ }\right)}{2^3\sin {20}^{\circ }}=\frac{1}{8}$

$\text{ So }\frac{\cos {20}^{\circ }+1}{{\sin }^2{80}^{\circ }}=\frac{2{\cos }^2{10}^{\circ }}{{\sin }^2{80}^{\circ }}=2$