(a) Start with the identity \(1 - \tan^2 \theta\).

We know \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), so \(1 - \tan^2 \theta = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\).

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

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

(b) Given \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} = 2 \tan^4 \theta\), substitute \(1 - \tan^2 \theta = 2 \tan^4 \theta\).

This simplifies to \(1 - \tan^2 \theta = 2 \tan^4 \theta \Rightarrow 2 \tan^4 \theta + \tan^2 \theta - 1 = 0\).

Let \(x = \tan^2 \theta\), then \(2x^2 + x - 1 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

\(x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\).

Thus, \(x = 0.5\) or \(x = -1\). Since \(x = \tan^2 \theta\), \(x = -1\) is not valid.

So, \(\tan^2 \theta = 0.5\) leading to \(\tan \theta = \pm \sqrt{0.5}\).

\(\theta = 35.3^\circ\) and \(\theta = 144.7^\circ\) (A.W.R.T).