9. Given $m, n, s \in \mathbf{R}^{+}, \alpha, \beta \in\left(0 \cdot \frac{\pi}{2}\right)$, and $m \tan \alpha+n \tan \beta=s$, then the minimum value of $m \sec \alpha$ $+n \sec \beta$ is $\qquad$
9. $\sqrt{(m+n)^{2}+s^{2}}$

Let $m \sec \alpha+n \sec \beta=y$, then $y^{2}=m^{2}\left(1+\tan ^{2} \alpha\right)+n^{2}\left(1+\tan ^{2} \beta\right)+$ $2 m n \sec \alpha \sec \beta=m^{2}+n^{2}+s^{2}+2 m n(\sec \alpha \sec \beta-\tan \alpha \tan \beta)$, since $\alpha, \beta \in$ $\left(0, \frac{\pi}{2}\right)$, then $\cos (\alpha-\beta) \leqslant 1$. That is, $\frac{1-\sin \alpha \sin \beta}{\cos \alpha \cos \beta} \geqslant 1$, which also means $\sec \alpha \sec \beta-\tan \alpha \tan \beta \geqslant 1$, so $y^{2} \geqslant(m+n)^{2}+s^{2}$, equality holds if and only if $\alpha=\beta=\arctan \frac{s}{m+n}$, the minimum value of $m \sec \alpha+n \sec \beta$ is $\sqrt{(m+n)^{2}+s^{2}}$.