(i) Express \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1}\) in the form \(a \sin^2 \theta + b\), where \(a\) and \(b\) are constants to be found.

(ii) Hence, or otherwise, and showing all necessary working, solve the equation \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = \frac{1}{4}\) for \(-90^\circ \leq \theta \leq 0^\circ\).