(a) (i) Prove that $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ - Leaving Cert Mathematics - Question 4 - 2022

First, we can express $\tan 15^\circ$ using known angles:

$\tan 15^\circ = \tan(45^\circ - 30^\circ)$

Using the formula for tan of a difference:

$\tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

Substituting values:

$\tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}}$

Thus:

$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Multiplying numerator and denominator by $\sqrt{3} - 1$ leads to:

$\tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

We can express $2 - \sqrt{3}$ as:

$\tan 15^\circ = \frac{\sqrt{3 - 1}}{\sqrt{3 + 1}}$

where $a = 3$.

Given that $|AB| = 10 \sqrt{2 - \sqrt{2}}$ and $|\angle ACB| = 45^\circ$, we can use the Law of Cosines.

First, denote the lengths:

$|AC| = |BC| = x$

By the Law of Cosines:

$|AB|^2 = |AC|^2 + |BC|^2 - 2 |AC||BC| \cos(45^\circ)$

Since $|AC| = |BC|$, we can simplify this:

$(10 \sqrt{2 - \sqrt{2}})^2 = x^2 + x^2 - 2x^2 \cdot \frac{\sqrt{2}}{2}$

This reduces to:

$100(2 - \sqrt{2}) = 2x^2(1 - \frac{\sqrt{2}}{2})$

Solving for $x$:

$100(2 - \sqrt{2}) = 2x^2 \cdot \frac{2 - \sqrt{2}}{2} \Rightarrow 100(2 - \sqrt{2}) = (2x^2 - \sqrt{2}x^2)$