$\mathrm{A}=\frac{\cos 9^0-\sin 9^0}{\cos 9^0+\sin 9^0}, \mathrm{B}=\frac{\cos 921+\sin {21}^0}{\cos {21}^0+\sin {21}^0}$ and 

$\mathrm{C} = \tan {20}^0 + \tan {40}^0 +\sqrt{3} \tan {20}^0 \tan {40}^0$ then

descending order of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is

$$\begin{gathered} \mathrm{A}=\frac{\cos 9^0-\sin 9^0}{\cos 9^0+\sin 9^0}=\tan ({45}^0-9^0)=\tan {36}^0 \\ \mathrm{B}=\frac{\cos 921+\sin {21}^0}{\cos {21}^0+\sin {21}^0}=\tan ({45}^0+{21}^0)=\tan {66}^0 \\ \mathrm{C}=\tan ({20}^0+{40}^0)=\tan {60}^0 \\ \frac{\tan {20}^0+\tan {40}^0}{1-\tan {20}^0 \tan {40}^0}=\sqrt{3} \\ \tan {20}^0+\tan {40}^0=\sqrt{3}-\sqrt{3} \tan {20}^0 \tan {40}^0 \\ \tan {20}^0+\tan {40}^0+\sqrt{3} \tan {20}^0 \tan {40}^0=\sqrt{3} \end{gathered}$$