3. As shown in Figure 2, quadrilaterals $A B C D, A_{1} B_{1} B A, \cdots$, $A_{100} B_{100} B_{99} A_{99}$ are all small squares with side length 1. Given $\angle A C B=\alpha, \angle A_{1} C B_{1}=\alpha_{1}, \cdots, \angle A_{100} C B_{100}$ $=\alpha_{100}$. Then $\tan \alpha \cdot \tan \alpha_{1}+\tan \alpha_{1} \cdot \tan \alpha_{2}+\cdots+$ $\tan \alpha_{99} \cdot \tan \alpha_{100}=(\quad)$.
(A) 1
(B) 100
(C) $\frac{99}{100}$
(D) $\frac{100}{101}$
3. D.
$$
\begin{array}{l}
\tan \alpha \cdot \tan \alpha_{1}+\tan \alpha_{1} \cdot \tan \alpha_{2}+\cdots+\tan \alpha_{99} \cdot \tan \alpha_{100} \\
=\frac{A B}{B C} \cdot \frac{A_{1} B_{1}}{B_{1} C}+\frac{A_{1} B_{1}}{B_{1} C} \cdot \frac{A_{2} B_{2}}{B_{2} C}+\cdots+\frac{A_{99} B_{99}}{B_{99} C} \cdot \frac{A_{100} B_{100}}{B_{100} C} \\
=\frac{1}{1} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{3}+\cdots+\frac{1}{100} \times \frac{1}{101} \\
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{100}-\frac{1}{101}=\frac{100}{101} .
\end{array}
$$