If $\mathrm{A}=\frac{2 \sin \mathrm{x}}{\sin 3 \mathrm{x}}+\frac{\tan \mathrm{x}}{\tan 3 \mathrm{x}},$

$\mathrm{B}=\cos {10}^0 \cos {30}^0 \cos {50}^0 \cos {70}^0,$

$\mathrm{C}=\tan {20}^0 \tan {40}^0 \tan {60}^0\tan {80}^0$

Then the ascending order of A, B, C is

$$\begin{gathered} \mathrm{A}=\frac{2 \sin \mathrm{x}}{\sin 3 \mathrm{x}}+\frac{\tan \mathrm{x}}{\tan 3 \mathrm{x}} \\ =\frac{\sin x}{\sin 3x}\left(2+\frac{\cos 3x}{\cos x}\right) \\ =\frac{\sin x}{\sin 3x}\left(\frac{\cos 3x+\cos x+\cos x}{\cos x}\right) \\ =\frac{\sin x}{\sin 3x}\frac{(2\cos 2x \cos x+\cos x)}{\cos x} \\ =\frac{4\sin x(2\cos 2x+1)}{3\sin x-2{\sin }^{3}x} \\ =\frac{4(3-4{\sin }^{2}x)}{3-4{\sin }^{2}x}=4 \end{gathered}$$
$$\begin{gathered} \mathrm{B}=\cos {10}^0 \cos {30}^0 \cos {50}^0 \cos {70}^0 \\ =\frac{\sqrt{3}}{2}\frac{1}{4}\cos {30}^o \\ =\frac{\sqrt{3}}{2}\frac{1}{4}\frac{\sqrt{3}}{2}=\frac{3}{16} \end{gathered}$$
$$\begin{gathered} \mathrm{C}=\tan {20}^0 \tan {40}^0 \tan {60}^0\tan {80}^0 \\ =\sqrt{3} \tan {60}^o \\ =\sqrt{3}\sqrt{3}=3 \end{gathered}$$