A geometry problem by KiXiao Cheng Leong

Note that the form of the two expressions multiplied is that of $4\cos^2 {\theta} -3$ , or $\dfrac{4\cos^3 {\theta} -3\cos{\theta}}{\cos{\theta}} = \dfrac{\cos{3\theta}}{\cos{\theta}}$ . Using this, we have

$(4\cos^2 {9^\circ}- 3)(4\cos^2 {27^\circ} - 3)$

$= \dfrac{(\cos{27^\circ})(\cos{81^\circ})}{(\cos{9^\circ})(\cos{27^\circ})}$

$= \dfrac{\cos{81^\circ}}{\cos{9^\circ}}$

$= \dfrac{\sin{9^\circ}}{\cos{9^\circ}}$

$= \boxed{ \tan{9^\circ}}$ .