曲线 $C$ 具有参数方程
$x = \mathrm{e}^t \cos t,\ y = \mathrm{e}^{-t} \sin t.$

1. Find the derivatives of $x$ and $y$ with respect to $t$ using the product rule.

$\frac{dx}{dt} = \mathrm{e}^t \cos t - \mathrm{e}^t \sin t = \mathrm{e}^t(\cos t - \sin t)$
$\frac{dy}{dt} = -\mathrm{e}^{-t} \sin t + \mathrm{e}^{-t} \cos t = \mathrm{e}^{-t}(\cos t - \sin t)$

2. Apply the chain rule to find the gradient function $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\mathrm{e}^{-t}(\cos t - \sin t)}{\mathrm{e}^t(\cos t - \sin t)} = \frac{\mathrm{e}^{-t}}{\mathrm{e}^t} = \mathrm{e}^{-2t}$

3. Determine the gradient and the coordinates of the point on the curve at parameter $t = p$.

Substitute $t = p$:
Gradient $m = \mathrm{e}^{-2p}$
Coordinates: $x_p = \mathrm{e}^p \cos p$ and $y_p = \mathrm{e}^{-p} \sin p$

4. Form the equation of the tangent and rearrange it into the form $y = mx + c$.

Using $y - y_p = m(x - x_p)$:
$y - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}(x - \mathrm{e}^p \cos p)$
$y - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}x - \mathrm{e}^{-p} \cos p$
$y = \mathrm{e}^{-2p}x - \mathrm{e}^{-p} \cos p + \mathrm{e}^{-p} \sin p$
$y = \mathrm{e}^{-2p}x + \mathrm{e}^{-p}(\sin p - \cos p)$

1. Find the equation of the tangent to the curve at $t = p$ (result required from part i).

$\frac{dx}{dt} = \mathrm{e}^t(\cos t - \sin t), \quad \frac{dy}{dt} = \mathrm{e}^{-t}(\cos t - \sin t)$
$\frac{dy}{dx} = \frac{\mathrm{e}^{-t}(\cos t - \sin t)}{\mathrm{e}^t(\cos t - \sin t)} = \mathrm{e}^{-2t}$
At $t=p$, the gradient is $m = \mathrm{e}^{-2p}$.
The equation of the tangent is:
$y - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}(x - \mathrm{e}^p \cos p)$
$y = \mathrm{e}^{-2p}x + \mathrm{e}^{-p}(\sin p - \cos p)$

2. Find the coordinates of $A$ and $B$ by evaluating the $x$ and $y$ intercepts of the tangent.

At the $y$-axis (point $B$), $x=0$:
$y_B = \mathrm{e}^{-p}(\sin p - \cos p)$
At the $x$-axis (point $A$), $y=0$:
$0 = \mathrm{e}^{-2p}x_A + \mathrm{e}^{-p}(\sin p - \cos p)$
$\mathrm{e}^{-2p}x_A = \mathrm{e}^{-p}(\cos p - \sin p)$
$x_A = \mathrm{e}^p(\cos p - \sin p)$

3. Formulate the area of triangle $OAB$, taking the absolute value to ensure a positive geometric area.

$\text{Area} = \frac{1}{2} |x_A| |y_B|$
$\text{Area} = \frac{1}{2} \left| \mathrm{e}^p(\cos p - \sin p) \right| \left| \mathrm{e}^{-p}(\sin p - \cos p) \right|$
$\text{Area} = \frac{1}{2} \mathrm{e}^p \mathrm{e}^{-p} \left| -(\cos p - \sin p)^2 \right|$
$\text{Area} = \frac{1}{2} (\cos p - \sin p)^2$

4. Simplify the area expression using trigonometric identities.

$\text{Area} = \frac{1}{2} (\cos^2 p - 2\sin p \cos p + \sin^2 p)$
$\text{Area} = \frac{1}{2} (1 - \sin 2p)$