$$
\text { 17. } \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \text { is }
$$

the value of $\qquad$ .
$\begin{array}{l}\text { 17. } \sqrt{3} \text {. } \\ \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \\ =\tan \left(24^{\circ}+36^{\circ}\right)\left(1-\tan 24^{\circ} \cdot \tan 36^{\circ}\right)+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ} \\ =\sqrt{3}-\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}=\sqrt{3} .\end{array}$