2015-II-18

Ans: A
Consider $\Delta ABC$, we have

$\begin{array}{rcl}
\tan \beta & = & \dfrac{AC}{AB} \\
AC & = & AB \tan \beta
\end{array}$

Consider $\Delta ACD$, we have

$\begin{array}{rcl}
\cos \alpha & = & \dfrac{AD}{AC} \\
\cos \alpha & = & \dfrac{AD}{AB \tan \beta} \\
\dfrac{AD}{AB} & = & \cos \alpha \tan \beta
\end{array}$