In triangle $ABC,$ right-angled at $B,$ if $\tan A = \frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C + \cos A \sin C$
$(ii)$ $\cos A \cos C - \sin A \sin C$

(N/A) Given $\tan A = \frac{1}{\sqrt{3}}.$
Since $\tan A = \frac{\text{Side opposite to } \angle A}{\text{Side adjacent to } \angle A} = \frac{BC}{AB},$
we have $\frac{BC}{AB} = \frac{1}{\sqrt{3}}.$
Let $BC = k$ and $AB = \sqrt{3}k,$ where $k$ is a positive real number.
In $\triangle ABC,$ by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (\sqrt{3}k)^2 + (k)^2 = 3k^2 + k^2 = 4k^2$
$AC = 2k$
Now,we find the trigonometric ratios:
$\sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$
$\cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$
$(i)$ $\sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
$(ii)$ $\cos A \cos C - \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$