Let's go step by step and prove both parts of the question.

Part (i):

We need to prove:

$\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)$

Using angle difference identities:

We know the following trigonometric identities for angle subtraction:

1. $\cos(x - y) = \cos(x)\cos(y) + \sin(x)\sin(y)$
2. $\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$

Now, applying the identity for $\cos(a - b)$:

$\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$

Similarly, for $\sin(a - b)$:

$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$

Now, substitute these identities into the left-hand side of the given equation:

$\cos(a)\cos(a - b) + \sin(a)\sin(a - b)$

Substitute $\cos(a - b)$ and $\sin(a - b)$ using the identities:

[ \cos(a)\left[\cos(a)\cos(b) + \sin(a)\sin(b)\right] + \sin(a)\left[\sin(a)\cos(b) - \cos(a)\sin(b)\right] ]

Now, expand this expression:

$= \cos(a)\cos(a)\cos(b) + \cos(a)\sin(a)\sin(b) + \sin(a)\sin(a)\cos(b) - \sin(a)\cos(a)\sin(b)$

Simplify the terms:

$= \cos^2(a)\cos(b) + \sin^2(a)\cos(b)$

Factor out $\cos(b)$:

$= \left(\cos^2(a) + \sin^2(a)\right)\cos(b)$

We know from the Pythagorean identity that $\cos^2(a) + \sin^2(a) = 1$. Therefore, the expression becomes:

$= 1 \cdot \cos(b)$

Hence, the left-hand side simplifies to:

$\cos(b)$

This proves that:

$\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)$

Part (ii):

We need to prove:

$\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)$

Using angle sum and difference identities:

We know the following identities for angle addition and subtraction:

1. $\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$
2. $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$

Now, multiply these two expressions:

$\cos(a + b)\cos(a - b) = \left(\cos(a)\cos(b) - \sin(a)\sin(b)\right) \left(\cos(a)\cos(b) + \sin(a)\sin(b)\right)$

This is in the form of $(x - y)(x + y)$, which simplifies to $x^2 - y^2$. So, we get:

$= \cos^2(a)\cos^2(b) - \sin^2(a)\sin^2(b)$

Now, if $\cos^2(b) = 1 - \sin^2(b)$, substitute this into the equation:

$= \cos^2(a)(1 - \sin^2(b)) - \sin^2(a)\sin^2(b)$

Expand the expression:

$= \cos^2(a) - \cos^2(a)\sin^2(b) - \sin^2(a)\sin^2(b)$

Thus, the equation simplifies to:

$= \cos^2(a) - \sin^2(b)$

Hence, we have proved that:

$\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)$

---

Summary:

1. For part (i), we showed that $\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)$.
2. For part (ii), we demonstrated that $\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)$.

---

Do you have any questions or would you like more details on any step?

Here are 5 related questions to expand on the topic:

1. What are the other trigonometric identities that can be derived from the angle sum and difference identities?
2. How can the identity for $\sin(a+b)$ be used in similar problems?
3. What is the relationship between the cosine of complementary angles?
4. How can the identity for $\cos(a + b)\cos(a - b)$ be applied in physics, particularly in wave interference?
5. Can these identities be extended to hyperbolic trigonometric functions?

Tip: Always try to use known identities in trigonometric proofs. Breaking down complex expressions using simple identities like the Pythagorean identity can simplify the process.