(i) Start with \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta} - 1}{\frac{\sin^2 \theta}{\cos^2 \theta} + 1}\).

Multiply numerator and denominator by \(\cos^2 \theta\):

\(\frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}\).

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to \(\sin^2 \theta - \cos^2 \theta = \sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta - 1\).

Thus, \(a = 2\) and \(b = -1\).

(ii) Solve \(2\sin^2 \theta - 1 = \frac{1}{4}\).

\(2\sin^2 \theta = \frac{1}{4} + 1 = \frac{5}{4}\).

\(\sin^2 \theta = \frac{5}{8}\).

\(\sin \theta = \pm \sqrt{\frac{5}{8}} \approx \pm 0.7906\).

Since \(-90^\circ \leq \theta \leq 0^\circ\), \(\theta = -52.2^\circ\).