Question 143, In a regular quadrilateral pyramid $P-ABCD$, $M$ and $N$ are the midpoints of $PA$ and $PB$ respectively, and the tangent value of the dihedral angle formed by the side face and the base is $\sqrt{2}$. Then, the cosine value of the angle formed by the skew lines $DM$ and $AN$ is $\qquad$.
Question 143, Solution Method One: As shown in the figure, let the center of the base $ABCD$ be O, and take the midpoint of $CD$ as K. Then $MN$ is parallel and equal to $DK$, so quadrilateral $MNKD$ is a parallelogram. Therefore, the angle formed by the skew lines $DM$ and $AN$ is $\angle ANK$. Let $AB=2$, then $OK=1, OD=\sqrt{2}, PO=\sqrt{2}$, hence $PD=2$, so $\triangle PAD$ is an equilateral triangle. Therefore, $KN=DM=AN=\sqrt{3}, AK=\sqrt{5}$. According to the cosine rule, $\cos \angle ANK = \frac{(\sqrt{3})^2 + (\sqrt{3})^2 - (\sqrt{5})^2}{2 \times \sqrt{3} \times \sqrt{3}} = \frac{1}{6}$.

Solution Method Two: Establish a spatial rectangular coordinate system with the center O of the base of the regular quadrilateral pyramid as the origin and $OP$ as the $z$-axis, such that $A(a, -a, 0), B(a, a, 0), C(-a, a, 0), D(-a, -a, 0)$. Then $P(0, 0, \sqrt{2}a)$, $M\left(\frac{a}{2}, -\frac{a}{2}, \frac{\sqrt{2}a}{2}\right), N\left(\frac{a}{2}, \frac{a}{2}, \frac{\sqrt{2}a}{2}\right)$, thus $\overrightarrow{DM} = \left(\frac{3a}{2}, \frac{a}{2}, \frac{\sqrt{2}a}{2}\right), \overrightarrow{AN} = \left(-\frac{a}{2}, \frac{3a}{2}, \frac{\sqrt{2}a}{2}\right)$. Therefore,