Prove that $\frac{2 - \sec^2 \theta}{\sec \theta (\sin \theta + \cos \theta)} = 1 - \tan \theta$. Hence solve the equation $\frac{4 - 2\sec^2 \theta}{\sec \theta (\sin \theta + \cos \theta)} = \sec^2 \theta - 2$ for $0 \le \theta \le 2\pi$.

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

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

2. Simplify the complex fraction by multiplying the numerator and denominator by $\cos^2 \theta$.

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

3. Use trigonometric identities to rewrite the numerator as a difference of squares.

Since $\cos 2\theta = 2\cos^2 \theta - 1 = \cos^2 \theta - \sin^2 \theta$,

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

4. Factorise the numerator, cancel the common factor, and simplify to reach the right-hand side (RHS).

$\text{LHS} = \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\cos \theta (\sin \theta + \cos \theta)}$

$= \frac{\cos \theta - \sin \theta}{\cos \theta}$

$= \frac{\cos \theta}{\cos \theta} - \frac{\sin \theta}{\cos \theta}$

$= 1 - \tan \theta = \text{RHS}$

1. Use the result from part (i) to simplify the left-hand side of the equation.

The left-hand side is $2 \times \frac{2 - \sec^2 \theta}{\sec \theta (\sin \theta + \cos \theta)}$.
Using (i), LHS $= 2(1 - \tan \theta)$.

2. Express the right-hand side in terms of $\tan \theta$.

RHS $= \sec^2 \theta - 2$
Since $\sec^2 \theta = 1 + \tan^2 \theta$,
RHS $= (1 + \tan^2 \theta) - 2 = \tan^2 \theta - 1$.

3. Form and solve the quadratic equation in $\tan \theta$.

$2(1 - \tan \theta) = \tan^2 \theta - 1$
$2 - 2\tan \theta = \tan^2 \theta - 1$
$\tan^2 \theta + 2\tan \theta - 3 = 0$
$(\tan \theta - 1)(\tan \theta + 3) = 0$
$\tan \theta = 1 \quad \text{or} \quad \tan \theta = -3$

4. Solve for $\theta$ when $\tan \theta = 1$.

For $\tan \theta = 1$ in $0 \le \theta \le 2\pi$:
$\theta = \frac{\pi}{4}, \pi + \frac{\pi}{4}$
$\theta = \frac{\pi}{4}, \frac{5\pi}{4}$

5. Solve for $\theta$ when $\tan \theta = -3$.

For $\tan \theta = -3$ in $0 \le \theta \le 2\pi$:
Basic angle $\alpha = \tan^{-1}(3) \approx 1.2490$
$\theta$ is in the 2nd and 4th quadrants.
$\theta = \pi - 1.2490 \approx 1.89$
$\theta = 2\pi - 1.2490 \approx 5.03$