Start with the left-hand side:

\(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = \frac{\cos \theta}{\frac{\sin \theta}{\cos \theta} (1 - \sin \theta)}\)

\(= \frac{\cos^2 \theta}{\sin \theta (1 - \sin \theta)}\)

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

\(= \frac{1 - \sin^2 \theta}{\sin \theta (1 - \sin \theta)}\)

Factor the numerator as a difference of squares:

\(= \frac{(1 - \sin \theta)(1 + \sin \theta)}{\sin \theta (1 - \sin \theta)}\)

Cancel \(1 - \sin \theta\) from the numerator and denominator:

\(= \frac{1 + \sin \theta}{\sin \theta}\)

Split the fraction:

\(= \frac{1}{\sin \theta} + 1\)

This matches the right-hand side, proving the identity.