Start with the left-hand side:

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

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

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

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

Factor the denominator:

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

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

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

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