3.10. In the square $A B C D$, an isosceles triangle $A E F$ is inscribed; point $E$ lies on side $B C$, point $F$ lies on side $C D$, and $A E=A F$. The tangent of angle $A E F$ is 3. Find the cosine of angle $F A D$.
3.10. Given: $ABCD-$ is a square, $AEF$ is an isosceles triangle, $AE=AF, E \in BC, F \in CD, \operatorname{tg} \angle AEF = 3$ (Fig. 3.12); we need to find $\cos \angle FAD$. Let $\angle AEF = \alpha, \angle FAD = \beta$; then $\angle EAF = 180^\circ - 2\alpha$. Since $\triangle ABE = \triangle ADF$ (by leg and hypotenuse), we have $\beta = \frac{1}{2}(90^\circ - 180^\circ + 2\alpha) = \alpha - 45^\circ$.

Therefore, $\cos \beta = \cos(\alpha - 45^\circ) = \frac{\sqrt{2}}{2}(\cos \alpha + \sin \alpha)$. Considering that $\operatorname{tg} \alpha = 3$, we find $1 + 9 = \frac{1}{\cos^2 \alpha}$, from which $\cos \alpha = \frac{1}{\sqrt{10}}, \sin \alpha = \operatorname{tg} \alpha \cos \alpha = \frac{3}{\sqrt{10}}$. Thus, $\cos \beta = \frac{\sqrt{2}}{2} \cdot \frac{4}{\sqrt{10}} = \frac{2\sqrt{5}}{5}$.