Tricky Trigonometry

Using the identity

$\begin{aligned} \cos 36^\circ + \cos 108^\circ & = \frac{1}{2} \\ \Rightarrow \cos 108^\circ & = \frac{1}{2} - \cos 36^\circ \\ & = \frac{1}{2} - \frac{1+\sqrt{5}}{4} \\ & = \frac{1-\sqrt{5}}{4} \end{aligned}$

Using the identity

$\begin{aligned} \cos 108^\circ & = \frac{1-\tan^2 54^\circ}{1+ \tan^2 54^\circ} \\ \Rightarrow \frac{1-\tan^2 54^\circ}{1+ \tan^2 54^\circ} & = \frac{1-\sqrt{5}}{4} \\ 4(1-\tan^2 54^\circ)& = (1-\sqrt{5})(1+ \tan^2 54^\circ ) \\ \Rightarrow \tan^2 54^\circ & = \frac {3+\sqrt{5}}{5-\sqrt{5}} \end{aligned}$

Similarly, we get: $\tan^2 18^\circ = \dfrac {3-\sqrt{5}}{5+\sqrt{5}}$

Therefore,

$\begin{aligned} \tan^2 18^\circ \tan^2 54^\circ& = \left(\dfrac {3-\sqrt{5}}{5+\sqrt{5}}\right) \left(\dfrac {3+\sqrt{5}}{5-\sqrt{5}} \right) \\ & = \frac{9-5}{25-5} = \frac{4}{20} = \frac{1}{5} = \boxed{0.2} \end{aligned}$