$\begin{aligned} &\left(4 \cos ^2 9^{\circ}-3\right)\left(4 \cos ^2 27^{\circ}-3\right) \\ &= \frac{1}{\cos 9^{\circ}}\left(4 \cos ^3 9^{\circ}-3 \cos 9^{\circ}\right) \times \frac{1}{\cos 27^{\circ}} \\ &\left(4 \cos ^3 27^{\circ}-3 \cos 27^{\circ}\right) \\ &= \frac{1}{\cos 9^{\circ}}\left(\cos 27^{\circ}\right) \times \frac{1}{\cos 27^{\circ}}\left(\cos 81^{\circ}\right) \\ &= \frac{\cos \left(90^{\circ}-9^{\circ}\right)}{\cos 9^{\circ}}=\frac{\sin 9^{\circ}}{\cos 9^{\circ}}=\tan 9^{\circ}\end{aligned}$