9. (i) (Grade 10) $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=$ $\qquad$ .
(ii) (Grade 11) Using the digits $1,2,3,4,5,6$ to number the six faces of a cube. If a numbered cube can be rotated to match the numbering of another cube, they are considered the same numbering method. Therefore, the number of different numbering methods is $\qquad$ kinds.
9. (i) $\sqrt{2}$.

Since $\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}$
$=\cos 40^{\circ} \sqrt{1-\cos 80^{\circ}}$
$=\cos 40^{\circ} \cdot \sqrt{2} \sin 40^{\circ}=\frac{\sqrt{2}}{2} \sin 80^{\circ}=\frac{\sqrt{2}}{2} \cos 10^{\circ}$,
therefore, $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=\sqrt{2}$.
(ii) 30.
Place the face marked with 6 on the table, the face opposite to 6 can be marked in 5 different ways, and the four middle faces can be marked with the remaining four numbers, considering the rotation of the cube around the vertical axis through the center of the base, there are 3! ways to mark the numbers. Therefore, the total number of different marking methods is $5 \times 3!=30$.