4. A.

Extend $A C$ to point $E$, such that $E C=A C$; connect $E B$, intersecting arc $\overparen{B C}$ at point $F$.
Since $A B$ is the diameter $\Rightarrow A C \perp B C$
$$
\Rightarrow A C=E C, A B=E B=10 \text {. }
$$

By symmetry, we know $E F=A D=4$.
By the secant theorem, we have
$$
\begin{array}{l}
E C \cdot E A=E F \cdot E B \\
\Rightarrow 2 A C^{2}=40 \Rightarrow A C=2 \sqrt{5} .
\end{array}
$$

By the Pythagorean theorem, we get $B C=4 \sqrt{5}$.
Thus $\tan B=\frac{A C}{B C}=\frac{1}{2}$.
4. As shown in Figure 2, a chord $BC$ (not a diameter) of the semicircle $\odot O$ is used as the axis of symmetry to fold the arc $\overparen{BC}$, intersecting the diameter at point $D$. If $AD=4, BD=6$, then $\tan B=$ ( .
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{4}{5}$