Given that the area of $\triangle ABC$ is $\frac{1}{4}$, if $\cos 2A + \cos 2B + 2\sin C = 2$, $\cos A \cos B \sin C = \frac{1}{4}$, then
A. $\sin C = \sin^2 A + \sin^2 B$
B. $AB = \sqrt{2}$
C. $\sin A + \sin B = \frac{\sqrt{6}}{2}$
D. $AC^2 + BC^2 = 3$
Given that the area of $\triangle ABC$ is $\frac{1}{4}$, if $\cos 2A + \cos 2B + 2\sin C = 2$, $\cos A \cos B \sin C = \frac{1}{4}$, then\\
A. $\sin C = \sin^2 A + \sin^2 B$\\
B. $AB = \sqrt{2}$\\
C. $\sin A + \sin B = \frac{\sqrt{6}}{2}$\\
D. $AC^2 + BC^2 = 3$