Show that $\cos(A + B) \cos(A - B) = \cos^2 A + \cos^2 B - 1$. Hence, determine the value of $\cos 15^\circ \cos 75^\circ$ without the use of calculator.

1. Expand the terms $\cos(A + B)$ and $\cos(A - B)$ using the addition formulae.

$\text{LHS} = (\cos A \cos B - \sin A \sin B)(\cos A \cos B + \sin A \sin B)$

2. Multiply the expanded terms using the difference of squares identity.

$\text{LHS} = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B$

3. Express $\sin^2 A$ and $\sin^2 B$ in terms of $\cos^2 A$ and $\cos^2 B$.

$\text{LHS} = \cos^2 A \cos^2 B - (1 - \cos^2 A)(1 - \cos^2 B)$

4. Expand and simplify to obtain the right-hand side.

$\text{LHS} = \cos^2 A \cos^2 B - (1 - \cos^2 B - \cos^2 A + \cos^2 A \cos^2 B)$
$= \cos^2 A \cos^2 B - 1 + \cos^2 B + \cos^2 A - \cos^2 A \cos^2 B$
$= \cos^2 A + \cos^2 B - 1$
$= \text{RHS} \text{ (Shown)}$

1. Identify suitable values for $A$ and $B$ to match the expression $\cos 75^\circ \cos 15^\circ$ with the identity from part (i).

Let $A + B = 75^\circ$ and $A - B = 15^\circ$.
Solving these simultaneous equations:
$2A = 90^\circ \implies A = 45^\circ$
$2B = 60^\circ \implies B = 30^\circ$

2. Substitute $A = 45^\circ$ and $B = 30^\circ$ into the identity from part (i).

Using $\cos(A + B) \cos(A - B) = \cos^2 A + \cos^2 B - 1$:
$\cos(45^\circ + 30^\circ) \cos(45^\circ - 30^\circ) = \cos^2 45^\circ + \cos^2 30^\circ - 1$
$\cos 75^\circ \cos 15^\circ = \cos^2 45^\circ + \cos^2 30^\circ - 1$

3. Evaluate the expression using exact values for special angles.

We know that $\cos 45^\circ = \frac{\sqrt{2}}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
$\cos 75^\circ \cos 15^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - 1$
$= \frac{2}{4} + \frac{3}{4} - 1$
$= \frac{5}{4} - 1$
$= \frac{1}{4}$