Example 5. If $A=20^{\circ}, B=25^{\circ}$, find $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.

 untranslated part: 
- $(1+\operatorname{tg} B)$ 的值.

Note: The untranslated part is kept as is because "tg" is a notation for the tangent function, and the phrase "的值" means "the value of" in this context. However, for a complete translation, it should read: 
- the value of $(1+\operatorname{tg} B)$. 

Thus, the fully translated text is:

Example 5. If $A=20^{\circ}, B=25^{\circ}$, find the value of $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
$$
\text{Solve } \begin{aligned}
& (1+\tan A) \cdot(1+\tan B) \\
= & \left(1+\tan 20^{\circ}\right) \cdot\left(1+\tan 25^{\circ}\right) \\
= & 1+\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ} .
\end{aligned}
$$

Using the tangent addition formula in reverse $1=\tan 45^{\circ}$
$$
\begin{array}{l}
= \frac{\tan 20^{\circ}+\tan 25^{\circ}}{1-\tan 20^{\circ} \cdot \tan 25^{\circ}} \text{, we get } \\
1-\tan 20^{\circ} \cdot \tan 25^{\circ}=\tan 20^{\circ}+\tan 25^{\circ} .
\end{array}
$$
$$
\begin{array}{l}
\text{Therefore, } \tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ}=1 \text{. } \\
\text{Thus, }(1+\tan A) \cdot(1+\tan B)=2 .
\end{array}
$$