Why is the period of $f$, $\pi$? I came across a problem, which asked to compute the period of the function $$f(x)=3^{\sec^2x-\tan^2 x}.$$
The answer provided was $\pi$.
I don't get how.
 A: Note that $\sec^2 x- \tan^2 x= 1$ for all $x$ except those in the form $x = \dfrac{\pi}{2} + \pi k$ (where $\sec x$ and $\tan x$ are undefined. Thus, $f(x) = 3^{\sec^2 x- \tan^2 x}  = 3^1 = 3$ for all $x$ except those in the form $x = \dfrac{\pi}{2} + \pi k$, which are evenly spaced every $\pi$ units. Thus, $f(x)$ has period $\pi$.
A: It is a constant function. It is technically undefined at the points JimmyK4542 mentions.  In order for the identity $f(x+p)=f(x)$ to hold, is is required that the domain $D$ of $f$ satisfies $D+p=D$ (where $D+p$ is defined as $\{x+p:x\in D\}$).  For the domain in question, this occurs whenever $p$ is an integer multiple of $\pi$.  Because the function is constant, this is the only requirement.
However, that is unsatisfying because the function is essentially the constant function $3$, which has no smallest period.  The singularities are removable, and in many contexts it is useful to remove them (after briefly recognizing and acknowledging that there are singularities to be removed).
