Solution:

Given expression:
\[
\frac{{\tan \theta \cdot \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}
\]

To prove:
\[
\frac{{\tan \theta \cdot \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \sec \theta \cdot \tan \theta
\]

Proof:

Step 1:
Start by simplifying the given expression:
\[
\frac{{\tan \theta \cdot \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \frac{{\frac{{\sin \theta}}{{\cos \theta}} \cdot \frac{1}{{\cos \theta}} - 1}}{{\frac{{\sin \theta}}{{\cos \theta}} - \frac{1}{{\cos \theta}} + 1}}
\]

Step 2:
Simplify further:
\[
= \frac{{\frac{{\sin \theta}}{{\cos^2 \theta}} - 1}}{{\frac{{\sin \theta - 1}}{{\cos \theta}}}}
\]

Step 3:
Combine the fractions:
\[
= \frac{{\sin \theta - \cos^2 \theta}}{{\sin \theta - \cos \theta}}
\]

Step 4:
Use trigonometric identities:
\[
= \frac{{1 - \cos^2 \theta}}{{1 - \cos \theta \cdot \sin \theta}} = \frac{{\sin^2 \theta}}{{\sin \theta \cdot (1 - \sin \theta)}} = \frac{{\sin \theta}}{{\cos \theta}} = \tan \theta
\]

Step 5:
Since \(\tan \theta = \frac{{\sin \theta}}{{\cos \theta}}\) and \(\sec \theta = \frac{1}{{\cos \theta}}\), we have:
\[
\sec \theta \cdot \tan \theta = \frac{1}{{\cos \theta}} \cdot \frac{{\sin \theta}}{{\cos \theta}} = \frac{{\sin \theta}}{{\cos^2 \theta}} = \tan \theta
\]

Conclusion:
Therefore, \(\frac{{\tan \theta \cdot \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \sec \theta \cdot \tan \theta\).