10. 如图，$AB$是$\odot O$的直径，弦$CD \perp AB$，$AC = 2\sqrt{2}$，$BC = 1$，求$\cos \angle DBA$的值.

10.$\because CD \perp AB$，$\therefore \overset{\frown}{AC} = \overset{\frown}{AD}$，$\angle DBA = \angle CBA$。
又$\because AB$是$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}$。
在${\rm Rt}\triangle ABC$中，$AC = 2\sqrt{2}$，$BC = 1$，
由勾股定理，得$AB = 3$，
$\therefore \cos\angle DBA = \cos\angle CBA = \frac{1}{3}$。

11. 如图，在$ Rt \bigtriangleup ABC$中，$\angle ACB = 90^{\circ}$，$CM$为$AB$边上的中线，$AN \perp CM$，交$BC$于点$N$.若$CM = 3$，$AN = 4$，求$\tan \angle CAN$的值.

11.证$\triangle CAN \sim \triangle CBA$，$\therefore \frac{AC}{BC} = \frac{AN}{BA} = \frac{CN}{AC}$。
又$\because CM$为$AB$边上的中线，$\therefore M$为$AB$的中点。
$\therefore AB = 2CM = 6$，$\therefore \tan\angle CAN = \frac{CN}{AC} = \frac{AN}{AB} = \frac{2}{3}$。

12. 如图，把$n$个边长为$1$的正方形拼接成一排.
(1) 填空：$\tan \angle BA_{1}C =$，$\tan \angle BA_{2}C =$.
(2) 分别求出$\tan \angle BA_{3}C$，$\tan \angle BA_{4}C$的值.
(3) 仔细比较(1)(2)的结果，找出规律，按照规律直接写出$\tan \angle BA_{n}C$的值(用含$n$的代数式表示).

12.(1)$\frac{1}{3}$ (2)$\frac{1}{7}$，$\frac{1}{13}$ (3)$\frac{1}{n^2 - n + 1}$

(1)$\frac{1}{3}$，$\frac{1}{7}$
(2)过点$C$作$CD\perp BA_{3}$于点$D$。由图可知，$BC=1$，$A_{3}C$水平距离为$2$，竖直距离为$1$，则$BA_{3}$的斜率为$\frac{1}{3}$，方程为$y=\frac{1}{3}x$。点$C(1,0)$到直线$BA_{3}$的距离$CD=\frac{|\frac{1}{3}×1 - 0|}{\sqrt{(\frac{1}{3})^2 + 1}}=\frac{1}{\sqrt{10}}$，$BA_{3}=\sqrt{3^2 + 1^2}=\sqrt{10}$，则$\sin\angle BA_{3}C=\frac{CD}{A_{3}C}=\frac{\frac{1}{\sqrt{10}}}{\sqrt{5}}=\frac{1}{5\sqrt{2}}$，$\cos\angle BA_{3}C=\frac{2\sqrt{2}}{5}$，所以$\tan\angle BA_{3}C=\frac{1}{13}$。
同理可得，$\tan\angle BA_{4}C=\frac{1}{21}$
(3)$\frac{1}{n^2 - n + 1}$