7.7 If $\angle A=20^{\circ}, \angle B=25^{\circ}$, then the value of $(1+\operatorname{tg} A)(1+\operatorname{tg} B)$ is
(A) $\sqrt{3}$.
(B) 2 .
(C) $1+\sqrt{2}$.
(D) $2(\operatorname{tg} A+\operatorname{tg} B)$.
(E) None of the above conclusions is correct.
(36th American High School Mathematics Examination, 1985)
[Solution] Since $\operatorname{tg}\left(20^{\circ}+25^{\circ}\right)=\operatorname{tg} 45^{\circ}=1$, and

then
$$
\begin{array}{c}
\operatorname{tg}\left(20^{\circ}+25^{\circ}\right)=\frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 25^{\circ}}{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}} \\
\operatorname{tg} 20^{\circ}+\operatorname{tg} 25^{\circ}=1-\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ} \\
1-\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}-\operatorname{tg} 20^{\circ}-\operatorname{tg} 25^{\circ}=0 .
\end{array}
$$

and $\left(1+\operatorname{tg} 20^{\circ}\right)\left(1+\operatorname{tg} 25^{\circ}\right)=1+\operatorname{tg} 20^{\circ}+\operatorname{tg} 25^{\circ}+\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}$,
$$
\therefore \quad 2-\left(1-\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}-\operatorname{tg} 20^{\circ}-\operatorname{tg} 25^{\circ}\right)=2,
$$

which means
$$
1+\operatorname{tg} 20^{\circ} \operatorname{tg} 25^{\circ}+\operatorname{tg} 20^{\circ}+\operatorname{tg} 25^{\circ}=2 \text {. }
$$

Therefore, the answer is $(B)$.