How to determine the positive and negative nature of $a\sec\theta-b\tan\theta$?

Given $a,b\in\mathbb R^+$ and $a\gt b$, find the minimum value of $a\sec\theta-b\tan\theta$.

My approach:

Let $$x=a\sec\theta-b\tan\theta\\\implies x+b\tan\theta=a\sec\theta\\\implies x^2+b^2\tan^2\theta+2bx\tan\theta=a^2(1+\tan^2\theta)\\\implies(b^2-a^2)\tan^2\theta+2bx\tan\theta+x^2-a^2=0$$

Since $\tan\theta\in\mathbb R$ $$\implies D\ge0 \\ \implies4b^2x^2-4(b^2-a^2)(x^2-a^2)\ge0 \\ \implies b^2x^2-(b^2x^2-a^2b^2-a^2x^2+a^4)\ge0\\ \implies a^2b^2+a^2x^2-a^4\ge0\\ \implies b^2+x^2-a^2\ge0\\\implies x^2\ge a^2-b^2\\\implies x\le-\sqrt{a^2-b^2} \text{ or }x\ge\sqrt{a^2-b^2}$$

This is where my doubt is. If we consider $x\le-\sqrt{a^2-b^2}$ to be true then the minimum value of $x$ is not defined. If we consider $x\ge\sqrt{a^2-b^2}$ to be true then the minimum value is $\sqrt{a^2-b^2}$.

In the given statement, is it implied anywhere that the given expression is positive?

Your working shows you that the range of the function is $\mathbb{R}$ excluding the set $\{x:-\sqrt{a^2-b^2}<x<\sqrt{a^2-b^2}\}$

The local minimum is $\sqrt{a^2-b^2}$ and the local maximum is $-\sqrt{a^2-b^2}$.