Concept:
Express everything in terms of \(\sin\theta\) and \(\cos\theta\), factor, and use \(\sin^2\theta + \cos^2\theta = 1\) along with the double-angle identity \(\sin 2\theta = 2\sin\theta\cos\theta\) to find the range.
Step 1: Rewrite the expression in sines and cosines.
Let
\[
E = (\cot^2\theta - \cos^2\theta)(\tan^2\theta - \sin^2\theta).
\]
Using \(\cot\theta = \frac{\cos\theta}{\sin\theta}\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\[
\cot^2\theta - \cos^2\theta = \frac{\cos^2\theta}{\sin^2\theta} - \cos^2\theta = \frac{\cos^2\theta - \cos^2\theta\sin^2\theta}{\sin^2\theta} = \frac{\cos^2\theta(1-\sin^2\theta)}{\sin^2\theta}.
\]
Since \(1-\sin^2\theta = \cos^2\theta\):
\[
\cot^2\theta - \cos^2\theta = \frac{\cos^4\theta}{\sin^2\theta}.
\]
Similarly,
\[
\tan^2\theta - \sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta = \frac{\sin^2\theta - \sin^2\theta\cos^2\theta}{\cos^2\theta} = \frac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta} = \frac{\sin^4\theta}{\cos^2\theta}.
\]
Step 2: Multiply and simplify.
\[
E = \left(\frac{\cos^4\theta}{\sin^2\theta}\right)\left(\frac{\sin^4\theta}{\cos^2\theta}\right) = \sin^2\theta \cos^2\theta.
\]
Using \(\sin 2\theta = 2\sin\theta\cos\theta\), we have \(\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2 2\theta\).
Step 3: Determine the range for \(0 < \theta < \frac{\pi}{2}\).
For \(0 < \theta < \frac{\pi}{2}\), \(0 < 2\theta < \pi\), so \(0 < \sin 2\theta \le 1\). Hence \(0 < \sin^2 2\theta \le 1\).
\[
0 < E \le \frac{1}{4}.
\]
Thus \(\alpha = 0\) and \(\beta = \frac{1}{4}\).
Step 4: Compute \(\beta - \alpha\) and write the final answer.
\[
\beta - \alpha = \frac{1}{4} - 0 = \frac{1}{4}.
\]
\[
\boxed{\frac14}
\]