$(\tan^2(18^\circ))(\tan^2(54^\circ))$ is a rational number

Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks!

Use the fact that

$$\tan^2{18^{\circ}} = \frac{1-\cos{36^{\circ}}}{1+\cos{36^{\circ}}} = 1-\frac{2}{5} \sqrt{5}$$

Then use the fact that

$$\tan^2{54^{\circ}} = \frac{1}{\tan^2{36^{\circ}}}$$

so that

$$\tan^2{18^{\circ}} \tan^2{54^{\circ}} = \frac{\tan^2{18^{\circ}}}{\tan^2{36^{\circ}}} = \frac{1}{4} (1 -\tan^2{18^{\circ}})^2 = \frac{1}{5}$$

$$\tan18^\circ\tan54^\circ=\frac{2\sin18^\circ\sin54^\circ}{2\cos18^\circ\cos54^\circ}$$ $$=\frac{\cos36^\circ-\cos72^\circ}{\cos36^\circ+\cos72^\circ}$$ (applying $\cos(A\pm B)$ formulae)

$$=\frac{\frac{\sqrt5+1}4-\frac{\sqrt5+1}4}{\frac{\sqrt5+1}4+\frac{\sqrt5+1}4}$$ as $$\cos 72^\circ=2\cos^236^\circ-1=\frac{\sqrt5-1}4$$

$$\implies \tan18^\circ\tan54^\circ=\sqrt5$$

Hint:

Use: $$\sin(3x) = 3 \sin (x) \cos ^2(x)-\sin ^3(x)$$ $$\cos(3x) = \cos ^3(x)-3 \sin ^2(x) \cos (x)$$ Then plug in $x=18^\circ$.

Use the definition of tan (sin/cos) along with half angle relation for cos (i.e. 18=36/2), sum of angles relation for cos (i.e. 54=36+18) and the fact that sin^2=1-cos^2 to get the value for tan^2. Then massage the numbers based on the value for cos(36deg). You should end up with a ration number.