(a) (i) Expand \((\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\sin \theta \cos \theta + \sin^2 \theta = 1\).

This simplifies to \(\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 1\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have \(2\sin \theta \cos \theta = 0\).

This implies \(\sin 2\theta = 0\), leading to \(2\theta = 0, \pi, 2\pi\).

Thus, \(\theta = 0, \frac{1}{2}\pi, \pi\).

(ii) Verify \(\cos \theta + \sin \theta = 1\) for \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\).

For \(\theta = 0\), \(\cos 0 + \sin 0 = 1 + 0 = 1\).

For \(\theta = \frac{1}{2}\pi\), \(\cos \frac{1}{2}\pi + \sin \frac{1}{2}\pi = 0 + 1 = 1\).

(b) Prove the identity:

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

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

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

(c) Solve \(\frac{\cos \theta + \sin \theta - 1}{1 - 2 \sin^2 \theta} = 2(\cos \theta + \sin \theta - 1)\).

Let \(k = \cos \theta + \sin \theta - 1\), then \(\frac{k}{1 - 2 \sin^2 \theta} = 2k\).

\(1 = 2(1 - 2 \sin^2 \theta)\), leading to \(\sin^2 \theta = \frac{1}{4}\).

Thus, \(\sin \theta = \pm \frac{1}{2}\), giving solutions \(\theta = \frac{1}{6}\pi, \frac{5}{6}\pi\).

Including \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\) from part (a)(ii), the solutions are \(0, \frac{1}{6}\pi, \frac{1}{2}\pi, \frac{5}{6}\pi\).