You can choose a different domain for $\theta$; for example, $\theta \in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2}) $.

As $\theta$ ranges from $\pi$ to $3\pi/2$, $\sec(\theta)$ decreases from $-1$ to $-\infty$ and $\tan(\theta)$ ranges from $0$ to $+\infty$.

In particular, the substitution $x = \sec(\theta)$ still covers the domain $x \in (-\infty, -1)$, but you have $\tan(\theta)$ remaining positive.

Incidentally, a point often overlooked in introductory courses is that when you antidifferentiate a function on a disconnected domain, each component of the domain gets is own distinct constant of integration.

So, if you want to speak in the full generality of covering both the $x > 1$ and $x < -1$ domains, you need to remember that each of those domains gets is own separate constant of integration.

You can choose a different domain for $\theta$; for example, $\theta \in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2}) $.

As $\theta$ ranges from $\pi$ to $3\pi/2$, $\sec(\theta)$ decreases from $-1$ to $-\infty$ and $\tan(\theta)$ increases from $0$ to $+\infty$.

In particular, on this domain for $\theta$ the substitution $x = \sec(\theta)$ still covers the domain $x \in (-\infty, -1)$, but you have $\tan(\theta)$ remaining positive.

Incidentally, a point often overlooked in introductory courses is that when you antidifferentiate a function on a disconnected domain, each component of the domain gets is own distinct constant of integration.

So, if you want to speak in the full generality of covering both the $x > 1$ and $x < -1$ domains, you need to remember that each of those domains gets is own separate constant of integration.