(a(i))[3]
By expanding $(\cos\theta + \sin\theta)^2$ first, find the three solutions of $(\cos\theta + \sin\theta)^2 = 1$ when $0 \leq \theta \leq \pi$.
(a(ii))[2]
Hence confirm that the only solutions of $\cos\theta + \sin\theta = 1$ for $0 \leq \theta \leq \pi$ are $0$ and $\frac{1}{2}\pi$.
(b)[3]
Prove the identity by showing that $$\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = \frac{\cos\theta + \sin\theta - 1}{1 - 2\sin^2\theta}.$$
(c)[3]
Using the results of (a)(ii) and (b), solve the equation $$\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = 2(\cos\theta + \sin\theta - 1)$$ for $0 \leq \theta \leq \pi$.