曲线 $C$ 的参数方程定义为： $x = a\tan t, \qquad y = a\sec^{2}t\sin t, \qquad -\frac{\pi}{4} < t < \frac{\pi}{4},$ 其中 $a$ 为正常数。

1. Apply the Pythagorean identity to the expression under the square root.

$\text{LHS} = \frac{\tan t}{\sqrt{1+\tan^{2}t}}$
$= \frac{\tan t}{\sqrt{\sec^{2}t}}$

2. Simplify the denominator based on the given domain and express the ratio in terms of sine and cosine.

For $-\frac{\pi}{4} < t < \frac{\pi}{4}$, $\sec t > 0$, so $\sqrt{\sec^{2}t} = \sec t$.
$= \frac{\tan t}{\sec t}$
$= \frac{\left(\frac{\sin t}{\cos t}\right)}{\left(\frac{1}{\cos t}\right)}$
$= \sin t = \text{RHS}$

1. Express $y$ entirely in terms of $\tan t$ using a trigonometric identity and the result from part (a).

Using the identity $\sec^2 t = 1 + \tan^2 t$, we rewrite the expression for $y$:
$y = a(1 + \tan^2 t)\sin t$

From part (a), substitute $\sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}}$:
$y = a(1 + \tan^2 t) \left( \frac{\tan t}{\sqrt{1 + \tan^2 t}} \right)$
$y = a\tan t \sqrt{1 + \tan^2 t}$

2. Substitute $x$ to eliminate $t$ and simplify to find the cartesian equation.

From $x = a\tan t$, we have $\tan t = \frac{x}{a}$. Substituting this into our equation for $y$:
$y = a \left( \frac{x}{a} \right) \sqrt{1 + \left( \frac{x}{a} \right)^2}$
$y = x \sqrt{1 + \frac{x^2}{a^2}}$
$y = x \sqrt{\frac{a^2 + x^2}{a^2}}$

Since $a$ is a positive constant, $\sqrt{a^2} = a$. Thus, simplifying gives:
$y = \frac{x}{a}\sqrt{a^2 + x^2}$

1. Determine the cartesian equation $f(x)$ from part (b) and its domain, which are required for this part.

From part (b), we know $x = a\tan t \implies \tan t = \frac{x}{a}$.
Thus, $1+\tan^2 t = 1+\frac{x^2}{a^2} = \frac{a^2+x^2}{a^2}$.
Using the result from (a),
$\sin t = \frac{x/a}{\sqrt{\frac{a^2+x^2}{a^2}}} = \frac{x}{\sqrt{a^2+x^2}}$
$y = a\sec^2 t \sin t = a\left(\frac{a^2+x^2}{a^2}\right)\left(\frac{x}{\sqrt{a^2+x^2}}\right) = \frac{x}{a}\sqrt{a^2+x^2}$
Since $-\frac{\pi}{4} < t < \frac{\pi}{4}$, the domain is $-a < x < a$.
Therefore, $f(x) = \frac{x}{a}\sqrt{a^2+x^2}$.

2. Show that $f(-x) = -f(x)$.

Substitute $-x$ into $f(x)$:
$f(-x) = \frac{(-x)}{a}\sqrt{a^2+(-x)^2}$
$f(-x) = -\frac{x}{a}\sqrt{a^2+x^2}$
$f(-x) = -f(x)$
This shows that $f(x)$ is an odd function.

3. Sketch the curve $C$.

The curve passes through the origin $(0,0)$.
Since $f(-x) = -f(x)$, the graph has rotational symmetry about the origin.
At the domain limits $x \to \pm a$:
When $x \to a$, $y \to \frac{a}{a}\sqrt{a^2+a^2} = a\sqrt{2}$.
When $x \to -a$, $y \to -a\sqrt{2}$.
Since $f'(x) = \frac{2x^2+a^2}{a\sqrt{x^2+a^2}} > 0$ and $f''(x) = \frac{x(2x^2+3a^2)}{a(x^2+a^2)^{3/2}}$, the curve is strictly increasing, concave downwards for $-a < x < 0$ and concave upwards for $0 < x < a$.
(Note: The sketch is a smooth curve passing through $(0,0)$, ending with open circles at $(-a, -a\sqrt{2})$ and $(a, a\sqrt{2}))$.

1. Derive the Cartesian equation $y=f(x)$ (from part b) needed for the integration.

From part (a), $\sin t = \dfrac{\tan t}{\sqrt{1+\tan^2 t}}$.
Given $x = a\tan t \implies \tan t = \dfrac{x}{a}$, so $\sin t = \dfrac{x/a}{\sqrt{1+(x/a)^2}} = \dfrac{x}{\sqrt{a^2+x^2}}$.
$y = a\sec^2 t \sin t = a\left(1+\tan^2 t\right)\sin t = a\left(1+\dfrac{x^2}{a^2}\right)\dfrac{x}{\sqrt{a^2+x^2}} = \dfrac{x}{a}\sqrt{a^2+x^2}$.
Thus, $f(x) = \dfrac{x}{a}\sqrt{a^2+x^2}$.

2. Set up the definite integral for the area using the function's odd symmetry.

From part (c), $f(-x) = -f(x)$, so the curve has odd symmetry about the origin. The region bounded by $x=-A$ and $x=A$ consists of two identical areas (one above and one below the $x$-axis).
$\text{Area} = \int_{-A}^{0} |f(x)|\,dx + \int_{0}^{A} |f(x)|\,dx = 2\int_{0}^{A} f(x)\,dx$
$\text{Area} = 2\int_{0}^{A} \dfrac{x}{a}\sqrt{a^2+x^2}\,dx$

3. Evaluate the definite integral using a standard integration rule.

$\text{Area} = \dfrac{1}{a} \int_{0}^{A} 2x(a^2+x^2)^{1/2}\,dx$
$= \dfrac{1}{a} \left[ \dfrac{(a^2+x^2)^{3/2}}{3/2} \right]_{0}^{A}$
$= \dfrac{2}{3a} \left( (a^2+A^2)^{3/2} - (a^2+0^2)^{3/2} \right)$
Since $a > 0$, $(a^2)^{3/2} = a^3$.
$\text{Area} = \dfrac{2}{3a}\left( (a^2+A^2)^{3/2} - a^3 \right)$