<s>[INST] How to simplify $\sec\tan^{-1}\pi$? How to simplify $\sec\tan^{-1}\pi$?
I've never dealt with inverse trigonometry before. Thanks in advance.
 [/INST] Here is a right triangle.

In any right triangle, by definition, $\tan x = \frac ba$ and $\sec x = \frac ca$.
Also, of course, $a^2+b^2 = c^2$.
Since $\tan x =\frac ba $, we have $x = \tan^{-1}\frac ba$ and then:  $$\begin{align}\sec x & =  \frac ca \\ \sec\left(\tan^{-1} \frac ba\right) & = \frac ca \\ & = \frac {\sqrt{a^2+b^2}}a.\end{align}$$
Now take $b=\pi$ and $a=1$ and we have $$
\sec\tan^{-1} \frac \pi1 =  \frac {\sqrt{1^2+\pi^2}}1.$$

Also, observe: since this works for any right triangle, we can also take $a=3, b=4$ and obtain $$\sec\tan^{-1} \frac 43 =  \frac 53$$
without actually knowing what $\tan^{-1} \frac43$ is, and without even being able to calculate it.  
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