Solve the simultaneous equations
$\begin{aligned} y &= x^2 - 6, \\ y &= 2x - 3. \end{aligned}$ Prove that $\frac{3 \cos \theta + \cos 2\theta - 1}{\cos^2 \theta + 2 \cos \theta} = 2 - \sec \theta$.

1. Form a single quadratic equation in $x$ by equating the expressions for $y$.

Substitute $y = 2x - 3$ into $y = x^2 - 6$:
$2x - 3 = x^2 - 6$
$x^2 - 2x - 3 = 0$

2. Solve the quadratic equation for $x$.

$(x - 3)(x + 1) = 0$
$x = 3 \quad \text{or} \quad x = -1$

3. Find the corresponding $y$-values.

When $x = 3$:
$y = 2(3) - 3 = 3$

When $x = -1$:
$y = 2(-1) - 3 = -5$

1. Substitute the double angle formula for $\cos 2\theta$ into the numerator.

$\text{LHS} = \frac{3 \cos \theta + (2 \cos^2 \theta - 1) - 1}{\cos^2 \theta + 2 \cos \theta}$

2. Simplify the numerator.

$= \frac{2 \cos^2 \theta + 3 \cos \theta - 2}{\cos^2 \theta + 2 \cos \theta}$

3. Factorise the numerator and the denominator.

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

4. Cancel the common factor and express in terms of secant to reach the RHS.

$= \frac{2 \cos \theta - 1}{\cos \theta} = 2 - \frac{1}{\cos \theta} = 2 - \sec \theta = \text{RHS}$