The variables $x$ and $y$ are related by $y = \frac{2x+8}{2x+1}$. When values of $x(1-y)$ are plotted against $y$, a straight line is obtained. The straight line intersects the vertical and horizontal axes at $A$ and $B$ respectively.

1. Rearrange the given equation to group terms involving $x$ together

$y = \frac{2x+8}{2x+1}$
$y(2x+1) = 2x+8$
$2xy + y = 2x + 8$
$2xy - 2x = 8 - y$

2. Factorise to isolate the expression $x(1-y)$

$2x(y - 1) = 8 - y$
$x(y - 1) = \frac{8 - y}{2}$
$x(1 - y) = \frac{y - 8}{2}$

3. Express the equation in the linear form $Y = mX + c$

$x(1 - y) = \frac{1}{2}y - 4$
This is in the form $Y = mX + c$ where $Y = x(1-y)$ and $X = y$.

4. Find the coordinates of the intercepts $A$ and $B$

For the vertical intercept $A$, $X = 0$ (i.e., $y = 0$):
$Y = \frac{1}{2}(0) - 4 = -4$
So, $A$ is $(0, -4)$.

For the horizontal intercept $B$, $Y = 0$ (i.e., $x(1-y) = 0$):
$0 = \frac{1}{2}y - 4$
$\frac{1}{2}y = 4 \implies y = 8$
So, $B$ is $(8, 0)$.

1. Relate $\tan \theta$ to the gradient of the straight line

The angle $\theta$ is the angle the straight line makes with the positive horizontal axis. Therefore, $\tan \theta$ is equal to the gradient of the line.

2. Calculate the gradient of the line using the linear form derived in part (i)

From the linear form $x(1-y) = \frac{1}{2}y - 4$, the gradient $m$ is $\frac{1}{2}$.
Alternatively, using the coordinates of $A(0, -4)$ and $B(8, 0)$ from part (i):
$\text{Gradient} = \frac{0 - (-4)}{8 - 0} = \frac{4}{8} = \frac{1}{2}$
Thus, $\tan \theta = \frac{1}{2}$.