Is $\sec^{-1}(\sec(\pi/2)) = \pi/2$?

Question

I think it shouldn't be defined as $\pi/2$ is not in the range of the function $\sec^{-1}(x)$

Wolfram confused me by giving the answer as $\pi/2$ : Link

But it mentions on another page that $\pi/2$ is not in the range of $\sec^{-1}(x)$: Link

What is the correct answer?

Answer 1

$\sec (\frac \pi 2)$ is not defined, so any expression that uses it is undefined. Alpha is pattern matching, using $\sec^{-1}(\sec(x))=x$, which is true for most $x$. It is often sloppy on cases like this.

Answer 2

Ross is correct. To expand, Wolfram is probably doing something like this:

It notices that you are calling $f^{-1}(f(x))$.

Assuming the inverse exists, AND that $f(x)$ is valid, the output is x. Wolfram is simply pattern matching, probably to save computation. So Wolfram is never checking whether or not $f(x)$ is valid.

Similarly, to give a little more evidence this is what is happening, this happens here: http://www.wolframalpha.com/input/?i=e%5Elog%280%29