(11 Let $\sec x+\tan x=\frac{22}{7}, \csc x+\cot x=\frac{m}{n}(m, n$ have no common divisor greater than 1), then $m+n=$ $\qquad$
11 Given $\sec x+\tan x=\frac{22}{7}$, we know $\frac{1+\sin x}{\cos x}=\frac{22}{7}$. By the componendo and dividendo rule, we have
$$
\frac{1+\sin x+\cos x}{1+\sin x-\cos x}=\frac{29}{15},
$$

Also, because
$$
\frac{1+\sin x+\cos x}{1+\sin x-\cos x}=\frac{2 \cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}=\cot \frac{x}{2},
$$

and
$$
\begin{aligned}
\csc x+\cot x & =\frac{1+\cos x}{\sin x} \\
& =\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\cot \frac{x}{2}
\end{aligned}
$$

Therefore, $\frac{m}{n}=\frac{29}{15}, m+n=44$.