$\tan 6^\circ \tan 54^\circ \tan 66^\circ \cot 18^\circ = ?$

Let 6°=x. As we know, tanx•tan(60°-x)•tan(60°+x)=tan3x So, tan18°•cot18°=1

See Trigonometric Table: multiples of 3 degrees mostly in radicals .

$\sqrt{-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}+7} \sqrt{1+\frac{2}{\sqrt{5}}} \sqrt{-2 \sqrt{5}+2 \sqrt{15-6 \sqrt{5}}+7} \sqrt{2 \sqrt{5}+5}$

$\sqrt{\left(1+\frac{2}{\sqrt{5}}\right) \left(2 \sqrt{5}+5\right) \left(-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}+7\right) \left(-2 \sqrt{5}+2 \sqrt{15-6 \sqrt{5}}+7\right)}$

$1$