10. (11th "Hope Cup" Invitational Competition Question) Let the maximum value of the function $\sqrt{\tan x-1}+\sqrt{3-\tan x}$ be $M$, and the minimum value be $N$. Then what is $\frac{M}{N}$?
10. From $1 \leqslant \tan x \leqslant 3$, it is easy to know $\sqrt{\tan x-1}+\sqrt{3-\tan x} \geqslant \sqrt{2}, N=\sqrt{2}$. Let $\vec{a}=(\sqrt{\tan x-1}, \sqrt{3-\tan x}), \vec{b}=(1,1)$, we get
$$
\begin{aligned}
2 & =(\tan x-1)+(3-\tan x) \\
& =|\dot{a}|^{2} \geqslant \frac{(\dot{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(\sqrt{\tan x-1}+\sqrt{3-\tan x})^{2}}{1^{2}+1^{2}}=\frac{(\sqrt{\tan x-1}+\sqrt{3-\tan x})^{2}}{2},
\end{aligned}
$$

we can know $\sqrt{\tan x-1}+\sqrt{3-\tan x} \leqslant 2$.
$M=2$, and $N=\sqrt{2}$, so $\frac{M}{N}=\frac{2}{\sqrt{2}}=\sqrt{2}$.