Question:

Step 1: Introduce a substitution using the key trigonometric identity. We use the trigonometric identity $\sec^2 x - \tan^2 x = 1$. This can be factored as $(\sec x - \tan x)(\sec x + \tan x) = 1$. Let $u = \sec x + \tan x$. From the identity, we can express $\sec x - \tan x$ in terms of $u$: $$ \sec x - \tan x = \frac{1}{\sec x + \tan x} = \frac{1}{u} $$ Step 2: Simplify the numerator of the function $f(x)$. The numerator of $f(x)$ is $\sec x + \tan x - 1$. Using our substitution $u = \sec x + \tan x$: $$ \text{Numerator} = u - 1 $$ Step 3: Simplify the denominator of the function $f(x)$. The denominator of $f(x)$ is $\tan x - \sec x + 1$. We can rewrite this by factoring out a negative sign from the $\tan x - \sec x$ terms: $$ \text{Denominator} = -(\sec x - \tan x) + 1 $$ Now, substitute the expression for $\sec x - \tan x = 1/u$ from Step 1: $$ \text{Denominator} = -\frac{1}{u} + 1 = 1 - \frac{1}{u} = \frac{u-1}{u} $$ Step 4: Substitute the simplified numerator and denominator back into $f(x)$ and simplify. Now, substitute the simplified numerator from Step 2 and the simplified denominator from Step 3 into the expression for $f(x)$: $$ f(x) = \frac{\text{Numerator}}{\text{Denominator}} = \frac{u-1}{\frac{u-1}{u}} $$ For $x \in (0, \pi/2)$, we know $\sec x > 1$ and $\tan x > 0$, so $u = \sec x + \tan x > 1$. This implies $u-1 > 0$, so we can safely cancel the $(u-1)$ term: $$ f(x) = u $$ Therefore, the function simplifies to $f(x) = \sec x + \tan x$. Step 5: Determine the range of $f(x)$ for the given domain. We need to find the range of $f(x) = \sec x + \tan x$ for $x \in (0, \pi/2)$. Let's analyze the behavior of $\sec x$ and $\tan x$ in this interval: - As $x$ approaches $0$ from the right ($x \to 0^+$): - $\cos x \to 1^-$, so $\sec x = 1/\cos x \to 1^+$. - $\tan x = \sin x / \cos x \to 0^+/1^- = 0^+$. - Thus, $f(x) = \sec x + \tan x \to 1 + 0 = 1$. - As $x$ approaches $\pi/2$ from the left ($x \to (\pi/2)^-$): - $\cos x \to 0^+$, so $\sec x = 1/\cos x \to \infty$. - $\tan x = \sin x / \cos x \to 1/0^+ = \infty$. - Thus, $f(x) = \sec x + \tan x \to \infty + \infty = \infty$. Both $\sec x$ and $\tan x$ are strictly increasing functions on the interval $(0, \pi/2)$. Their sum $f(x)$ is also strictly increasing on this interval. Therefore, the range of $f(x)$ is $(1, \infty)$. The final answer is $\boxed{(1, \infty)}$.