Answer: 44

Small Hint:

If $u=\sec x+\tan x$, then $\sec x-\tan x=\frac{1}{u}$

Big Hint:

Rewrite $\csc x+\cot x$ as $\frac{\sec x+1}{\tan x}$

Solution:

Let $u=\sec x+\tan x=\frac{22}{7}.$ Since $\sec x+\tan x$ and $\sec x-\tan x$ have product $1,$ $$\begin{aligned}\sec x&=\frac{u+u^{-1}}2,\\\tan x&=\frac{u-u^{-1}}2.\end{aligned}$$ Also $$\begin{aligned}\csc x+\cot x&=\frac{1+\cos x}{\sin x}\\&=\frac{\sec x+1}{\tan x}\\&=\frac{u+1}{u-1}.\end{aligned}$$ Substituting $u=\frac{22}{7}$ gives $\frac{29}{15}.$ Thus $m+n=29+15=44.$