Given that $\theta$ is small, show that $\sec \theta \approx 1 + \tfrac{1}{2}\theta^2$. The diagram below shows a circle, centre $O$ and radius $r$, with points $A$ and $B$ on the circumference such that $\angle AOB = \theta$ radians, where $\theta$ is small. $AC$ is a tangent to the circle at $A$ and $OBC$ is a straight line. Show that the length of chord AB can be approximated by $r\theta$. Hence show that the perimeter of triangle ABC can be approximated by $r\theta(a + b\theta)$, where $a$ and $b$ are constants to be determined.

1. Express $\sec \theta$ in terms of $\cos \theta$ and substitute the small angle approximation for $\cos \theta$.

$\sec \theta = \frac{1}{\cos \theta}$

Since $\theta$ is small, we use the small angle approximation $\cos \theta \approx 1 - \frac{1}{2}\theta^2$.

Therefore, $\sec \theta \approx \frac{1}{1 - \frac{1}{2}\theta^2} = \left(1 - \frac{1}{2}\theta^2\right)^{-1}$.

2. Use the binomial expansion to expand the expression, ignoring terms in $\theta^4$ and higher.

Using the standard binomial series expansion $(1+x)^n \approx 1 + nx$ for small $x$:

$\left(1 - \frac{1}{2}\theta^2\right)^{-1} \approx 1 + (-1)\left(-\frac{1}{2}\theta^2\right)$

$\sec \theta \approx 1 + \frac{1}{2}\theta^2$ (Shown)

1. Formulate an expression for $AB^2$ using the cosine rule in $\triangle AOB$.

Using the cosine rule in $\triangle AOB$:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos \theta$
$AB^2 = r^2 + r^2 - 2(r)(r)\cos \theta$
$AB^2 = 2r^2(1 - \cos \theta)$

2. Apply the small angle approximation for cosine to demonstrate the approximation for $AB$.

For small $\theta$, we use the standard approximation $\cos \theta \approx 1 - \frac{1}{2}\theta^2$:
$AB^2 \approx 2r^2\left(1 - \left(1 - \frac{1}{2}\theta^2\right)\right)$
$AB^2 \approx 2r^2\left(\frac{1}{2}\theta^2\right) = r^2\theta^2$
Taking the positive square root (since $AB$, $r$, and $\theta$ are positive lengths/measures):
$AB \approx r\theta$

1. Express the lengths of $AC$ and $BC$ in terms of $r$ and $\theta$ using right-angled triangle $OAC$.

Since $AC$ is a tangent to the circle at $A$, $\angle OAC = 90^\circ$. In right-angled $\triangle OAC$,
$AC = OA \tan \theta = r\tan\theta$
$OC = OA \sec \theta = r\sec\theta$
Since $O$, $B$, and $C$ are collinear and $B$ lies on the circumference, $OB = r$. Thus,
$BC = OC - OB = r\sec\theta - r$

2. Apply small angle approximations to find $AC$ and $BC$ in terms of powers of $\theta$.

For small $\theta$, we use the standard approximation $\tan\theta \approx \theta$, so:
$AC \approx r\theta$
Using the result from part (a) that $\sec\theta \approx 1 + \frac{1}{2}\theta^2$:
$BC \approx r\left(1 + \frac{1}{2}\theta^2\right) - r = \frac{1}{2}r\theta^2$

3. Calculate the perimeter of triangle $ABC$ using the approximated side lengths.

The perimeter of $\triangle ABC$ is the sum of its sides $AB + BC + AC$.
Using the result from (b)(i) that $AB \approx r\theta$:
$\text{Perimeter} \approx r\theta + \frac{1}{2}r\theta^2 + r\theta$
$\text{Perimeter} \approx 2r\theta + \frac{1}{2}r\theta^2$

4. Factorise the perimeter expression to match the required format and state the constants.

Factoring out $r\theta$ from the approximation:
$\text{Perimeter} \approx r\theta\left(2 + \frac{1}{2}\theta\right)$
Comparing this with $r\theta(a + b\theta)$, we deduce that:
$a = 2$
$b = \frac{1}{2}$