Prove that $(\csc\theta + \cot\theta)^2 = \frac{1 + \cos\theta}{1 - \cos\theta}$. Hence, solve the equation $(\csc 2\theta + \cot 2\theta)^2 (1 - \cos 2\theta) = 2 \cos \theta$ for $0^\circ \le \theta \le 360^\circ$.

1. Express the left-hand side in terms of sine and cosine.

$\text{LHS} = (\csc\theta + \cot\theta)^2$
$\text{LHS} = \left(\frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta}\right)^2$

2. Combine the fractions and expand the square.

$\text{LHS} = \left(\frac{1 + \cos\theta}{\sin\theta}\right)^2$
$\text{LHS} = \frac{(1 + \cos\theta)^2}{\sin^2\theta}$

3. Apply the Pythagorean identity to the denominator.

$\text{LHS} = \frac{(1 + \cos\theta)^2}{1 - \cos^2\theta}$

4. Factorise the denominator and simplify to reach the right-hand side.

$\text{LHS} = \frac{(1 + \cos\theta)^2}{(1 - \cos\theta)(1 + \cos\theta)}$
$\text{LHS} = \frac{1 + \cos\theta}{1 - \cos\theta} = \text{RHS}$

1. Apply the result of (a) to simplify the left-hand side

From part (a), with $\theta$ replaced by $2\theta$:
$(\csc 2\theta + \cot 2\theta)^2 = \frac{1 + \cos 2\theta}{1 - \cos 2\theta}.$
Therefore
$(\csc 2\theta + \cot 2\theta)^2 (1 - \cos 2\theta) = \frac{1 + \cos 2\theta}{1 - \cos 2\theta}\times(1-\cos 2\theta) = 1 + \cos 2\theta.$

2. Use the double-angle formula and form the equation

Since $1 + \cos 2\theta = 2\cos^2\theta$, the equation becomes
$2\cos^2\theta = 2\cos\theta.$

3. Rearrange and factorise

$2\cos^2\theta - 2\cos\theta = 0$
$2\cos\theta(\cos\theta - 1) = 0$
So $\cos\theta = 0$ or $\cos\theta = 1$.

4. Solve for all angles in the interval

For $0^\circ \le \theta \le 360^\circ$:

$\cos\theta = 0 \Rightarrow \theta = 90^\circ,\ 270^\circ$

$\cos\theta = 1 \Rightarrow \theta = 0^\circ,\ 360^\circ$

Hence $\theta = 0^\circ,\ 90^\circ,\ 270^\circ,\ 360^\circ$.