Trigonometry + Radicals

$S=\sqrt{\dfrac{\sin 54^\circ}{\sin 18^\circ}}-\sqrt{\dfrac{\cos 72^\circ}{\cos 36^\circ}}$

$S=\sqrt{\dfrac{\cos 36^\circ}{\cos 72^\circ}}-\sqrt{\dfrac{\cos 72^\circ}{\cos 36^\circ}}$

$S=\dfrac{\cos 36^\circ-\cos 72^\circ}{\sqrt{\cos 36^\circ \cos 72^\circ}}$

Now use the product to sum formula in the denominator:

$S=\dfrac{-\cos 72^\circ - \cos 144^\circ}{\sqrt{\frac{1}{2}(\cos 108^\circ + \cos 36^\circ)}}$

$S=\sqrt{2}\dfrac{-\cos 72^\circ - \cos 144^\circ}{\sqrt{-\cos 72^\circ-\cos 144^\circ}}$

$S=\sqrt{2(-\cos 72^\circ-\cos 144^\circ)}$

Finally, let $w=e^{2 \pi i/5}$ , then we know that $w^4+w^3+w^2+w+1=0$ or $w^2+1/w^2+w+1/w+1=0$ , also $w^k+1/w^k=2\cos(72^\circ k)$ , so $\cos 144^\circ+\cos 72^\circ=-\dfrac{1}{2}$ , and $S=1$ .

$\begin{aligned}\\ \sqrt{\dfrac{Sin54}{Sin18}} -\sqrt{\dfrac{Cos72}{Cos36}}& =\sqrt{\dfrac{Sin54}{Sin18}}-\sqrt{\dfrac{Sin18}{Sin54}}\\ &=\sqrt{\dfrac{3Sin18-4(Sin18)^3}{Sin18}}-\sqrt{\dfrac{Sin18}{3Sin18-4(Sin18)^3}}\\ &=\sqrt{3-4(Sin18)^2}-\sqrt{\dfrac 1 {3-4(Sin18)^2}}\\ &=\sqrt{3-\dfrac{3-\sqrt5} 2} - \sqrt{\dfrac 1 {3-\dfrac{3-\sqrt5} 2}} \\ &=\sqrt{ \dfrac{3+\sqrt5} 2} - \sqrt{ \dfrac {3-\sqrt5} 2} \\ &=\Large~~\color{#D61F06}{1}.\\ \end{aligned} \\$