A curve C is defined by the parametric equations $x = \cos\theta - \dfrac{1}{\cos\theta},\ \ y = \cos\theta + \dfrac{1}{\cos\theta},\ \ -\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}.$ The line L is the normal to C at the point with parameter $p$.

1. Differentiate $x$ with respect to $\theta$.

$x = \cos\theta - (\cos\theta)^{-1}$
$\frac{dx}{d\theta} = -\sin\theta - (-1)(\cos\theta)^{-2}(-\sin\theta)$
$\frac{dx}{d\theta} = -\sin\theta - \frac{\sin\theta}{\cos^{2}\theta}$
$\frac{dx}{d\theta} = -\sin\theta \left( 1 + \frac{1}{\cos^{2}\theta} \right) = -\sin\theta \left( \frac{\cos^{2}\theta + 1}{\cos^{2}\theta} \right)$

2. Differentiate $y$ with respect to $\theta$.

$y = \cos\theta + (\cos\theta)^{-1}$
$\frac{dy}{d\theta} = -\sin\theta + (-1)(\cos\theta)^{-2}(-\sin\theta)$
$\frac{dy}{d\theta} = -\sin\theta + \frac{\sin\theta}{\cos^{2}\theta}$
$\frac{dy}{d\theta} = -\sin\theta \left( 1 - \frac{1}{\cos^{2}\theta} \right) = -\sin\theta \left( \frac{\cos^{2}\theta - 1}{\cos^{2}\theta} \right)$

3. Apply the chain rule for parametric equations to find $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
$\frac{dy}{dx} = \frac{-\sin\theta \left( \frac{\cos^{2}\theta - 1}{\cos^{2}\theta} \right)}{-\sin\theta \left( \frac{\cos^{2}\theta + 1}{\cos^{2}\theta} \right)}$
Cancelling the common factor $-\sin\theta$ and the common denominator $\cos^{2}\theta$ yields:
$\frac{dy}{dx} = \frac{\cos^{2}\theta - 1}{\cos^{2}\theta + 1} \quad \text{(Shown)}$

1. Determine the gradient of the normal L at parameter $p$.

From (i), the gradient of the tangent at parameter $p$ is $m_t = \dfrac{\cos^2 p - 1}{\cos^2 p + 1}$.
The gradient of the normal L is the negative reciprocal:
$m_n = -\dfrac{1}{m_t} = -\dfrac{\cos^2 p + 1}{\cos^2 p - 1} = \dfrac{1 + \cos^2 p}{1 - \cos^2 p}$.

2. Set up the equation of the normal L using the coordinates of the point at parameter $p$.

At parameter $p$, the coordinates are:
$x = \cos p - \dfrac{1}{\cos p} = \dfrac{\cos^2 p - 1}{\cos p}$
$y = \cos p + \dfrac{1}{\cos p} = \dfrac{\cos^2 p + 1}{\cos p}$

Using the point-gradient form $y - y_1 = m_n(x - x_1)$, the equation of the normal is:
$y - \left(\dfrac{\cos^2 p + 1}{\cos p}\right) = \dfrac{1 + \cos^2 p}{1 - \cos^2 p} \left[ x - \left(\dfrac{\cos^2 p - 1}{\cos p}\right) \right]$

3. Simplify the equation to match the required form.

Expanding the right-hand side and moving the $y$-coordinate term over:
$y = \dfrac{1 + \cos^2 p}{1 - \cos^2 p}x - \left(\dfrac{1 + \cos^2 p}{1 - \cos^2 p}\right) \left(\dfrac{\cos^2 p - 1}{\cos p}\right) + \dfrac{\cos^2 p + 1}{\cos p}$

Notice that $\dfrac{\cos^2 p - 1}{1 - \cos^2 p} = -1$, so the equation simplifies to:
$y = \dfrac{1 + \cos^2 p}{1 - \cos^2 p}x - (1 + \cos^2 p)\left(-\dfrac{1}{\cos p}\right) + \dfrac{1 + \cos^2 p}{\cos p}$
$y = \dfrac{1 + \cos^2 p}{1 - \cos^2 p}x + \dfrac{1 + \cos^2 p}{\cos p} + \dfrac{1 + \cos^2 p}{\cos p}$
$y = \dfrac{1 + \cos^2 p}{1 - \cos^2 p}x + 2\left(\dfrac{1 + \cos^2 p}{\cos p}\right) \quad \text{(shown)}$

1. Find the $y$-intercept $Q$ by setting $x = 0$ in the equation of line $L$.

$y = \frac{1+\cos^{2}p}{1-\cos^{2}p}(0) + 2\left(\frac{1+\cos^{2}p}{\cos p}\right)$
$y_Q = \frac{2(1+\cos^{2}p)}{\cos p}$
Thus, the $y$-intercept $Q$ is at $\left(0, \frac{2(1+\cos^{2}p)}{\cos p}\right)$.

2. Find the $x$-intercept $P$ by setting $y = 0$ in the equation of line $L$.

$0 = \frac{1+\cos^{2}p}{1-\cos^{2}p}\,x + 2\left(\frac{1+\cos^{2}p}{\cos p}\right)$
$\frac{1+\cos^{2}p}{1-\cos^{2}p}\,x = - \frac{2(1+\cos^{2}p)}{\cos p}$
Dividing both sides by $1+\cos^{2}p$ (which is strictly positive and therefore non-zero):
$\frac{x}{1-\cos^{2}p} = - \frac{2}{\cos p}$
$x_P = - \frac{2(1-\cos^{2}p)}{\cos p}$
Thus, the $x$-intercept $P$ is at $\left(- \frac{2(1-\cos^{2}p)}{\cos p}, 0\right)$.

3. Calculate the area of the triangle $OPQ$.

The area of right-angled triangle $OPQ$ is given by $\frac{1}{2} |x_P| |y_Q|$.
Since $-\frac{\pi}{2} < p < \frac{\pi}{2}$, we have $\cos p > 0$.
Thus $1-\cos^2 p \ge 0$, meaning $x_P \le 0$ and $y_Q \ge 0$.
$\text{Area} = \frac{1}{2} \left| - \frac{2(1-\cos^{2}p)}{\cos p} \right| \left| \frac{2(1+\cos^{2}p)}{\cos p} \right|$
$\text{Area} = \frac{1}{2} \left( \frac{2(1-\cos^{2}p)}{\cos p} \right) \left( \frac{2(1+\cos^{2}p)}{\cos p} \right)$
$\text{Area} = \frac{2(1-\cos^{2}p)(1+\cos^{2}p)}{\cos^{2}p}$
$\text{Area} = \frac{2(1-\cos^{4}p)}{\cos^{2}p}$
(Note: This can equivalently be written as $2\sec^2 p - 2\cos^2 p$ or $2\tan^2 p (1+\cos^2 p)$)

1. Establish the expression for the area of triangle OPQ by determining the intercepts of the normal line $L$ (from part ii).

From part (ii), the normal line $L$ is given by $y = \frac{1+\cos^{2}p}{1-\cos^{2}p}x + 2\left(\frac{1+\cos^{2}p}{\cos p}\right)$.
To find the area required from part (iii), we determine the $x$ and $y$ intercepts of $L$.
Setting $y = 0$, the $x$-intercept is:
$x_P = -2\left(\frac{1+\cos^{2}p}{\cos p}\right) \div \left(\frac{1+\cos^{2}p}{1-\cos^{2}p}\right) = -\frac{2(1-\cos^2 p)}{\cos p} = -\frac{2\sin^2 p}{\cos p}$
Setting $x = 0$, the $y$-intercept is:
$y_Q = \frac{2(1+\cos^{2}p)}{\cos p}$
Since $-\frac{\pi}{2} < p < \frac{\pi}{2}$, $\cos p > 0$. The area $A$ of triangle OPQ is:
$A = \frac{1}{2} |x_P y_Q| = \frac{1}{2} \left( \frac{2\sin^2 p}{\cos p} \right) \left( \frac{2(1+\cos^2 p)}{\cos p} \right)$
$A = \frac{2\sin^2 p(1+\cos^2 p)}{\cos^2 p} = 2\tan^2 p(1+\cos^2 p)$
Expanding this gives a simpler form for differentiation:
$A = 2\tan^2 p + 2\sin^2 p$

2. Differentiate the area $A$ with respect to the parameter $p$.

Differentiate the area $A$ with respect to $p$:
$\frac{\mathrm{d}A}{\mathrm{d}p} = \frac{\mathrm{d}}{\mathrm{d}p}(2\tan^2 p + 2\sin^2 p)$
Using the chain rule for each trigonometric term:
$\frac{\mathrm{d}A}{\mathrm{d}p} = 2(2\tan p \sec^2 p) + 2(2\sin p \cos p)$
$\frac{\mathrm{d}A}{\mathrm{d}p} = 4\tan p \sec^2 p + 4\sin p \cos p$

3. Evaluate the derivative $\frac{\mathrm{d}A}{\mathrm{d}p}$ at the specific value $p = \frac{\pi}{3}$.

Substitute $p = \frac{\pi}{3}$ into the derivative. Note the exact values:
$\tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \sec\left(\frac{\pi}{3}\right) = 2, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$
Substituting these into our expression:
$\frac{\mathrm{d}A}{\mathrm{d}p} \bigg|_{p=\frac{\pi}{3}} = 4(\sqrt{3})(2)^2 + 4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
$\frac{\mathrm{d}A}{\mathrm{d}p} \bigg|_{p=\frac{\pi}{3}} = 16\sqrt{3} + \sqrt{3} = 17\sqrt{3}$

4. Apply the chain rule for connected rates of change to find the rate of increase of the area.

Using the connected rates of change formula:
$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}p} \times \frac{\mathrm{d}p}{\mathrm{d}t}$
Given that $p$ increases at a constant rate of $0.1$ units per second ($\frac{\mathrm{d}p}{\mathrm{d}t} = 0.1$):
$\frac{\mathrm{d}A}{\mathrm{d}t} = 17\sqrt{3} \times 0.1 = 1.7\sqrt{3}$