Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms.  Find $m+n^{}_{}.$
Use the two [trigonometric Pythagorean identities](https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Pythagorean_Identities) $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$. 
If we square the given $\sec x = \frac{22}{7} - \tan x$, we find that 
\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}
This yields $\tan x = \frac{435}{308}$. 
Let $y = \frac mn$. Then squaring, 
\[\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.\]
Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation): $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$. It turns out that only the [positive](https://artofproblemsolving.com/wiki/index.php/Positive) root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$.
Note: The problem is much easier computed if we consider what $\sec (x)$ is, then find the relationship between $\sin( x)$ and $cos (x)$ (using $\tan (x) = \frac{435}{308}$, and then computing $\csc x + \cot x$ using $1/\sin x$ and then the reciprocal of $\tan x$.