(i) Start with the left-hand side: \(\tan^2 \theta - \sin^2 \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

Thus, \(\tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta\).

Factor out \(\sin^2 \theta\):

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

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

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

Thus, \(\tan^2 \theta - \sin^2 \theta \equiv \tan^2 \theta \sin^2 \theta\).

(ii) From the identity, \(\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta\), the right-hand side is positive for \(0^\circ < \theta < 90^\circ\) because \(\tan^2 \theta\) and \(\sin^2 \theta\) are positive.

Therefore, \(\tan^2 \theta > \sin^2 \theta\), implying \(\tan \theta > \sin \theta\) for \(0^\circ < \theta < 90^\circ\).