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

1. Apply the Pythagorean trigonometric identity to the denominator.

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

2. Simplify the square root using the given domain of $t$, and express in terms of sine and cosine to complete the proof.

$\begin{aligned} \text{LHS} &= \frac{\tan t}{\sec t} \quad \left(\text{since } \sec t > 0 \text{ for } -\frac{\pi}{4} < t < \frac{\pi}{4}\right) \\ &= \frac{\left(\frac{\sin t}{\cos t}\right)}{\left(\frac{1}{\cos t}\right)} \\ &= \sin t \\ &= \text{RHS} \end{aligned}$

1. Express $y$ entirely in terms of $\tan t$ by using the identity $\sec^2 t = 1 + \tan^2 t$ and the result from part (a).

We are given $y = a \sec^2 t \sin t$.
Using the identity $\sec^2 t = 1 + \tan^2 t$ and substituting $\sin t = \dfrac{\tan t}{\sqrt{1 + \tan^2 t}}$ from part (a):
$y = a(1 + \tan^2 t) \left( \frac{\tan t}{\sqrt{1 + \tan^2 t}} \right)$
$y = a \tan t \sqrt{1 + \tan^2 t}$

2. Substitute $\tan t = \frac{x}{a}$ into the expression and simplify to obtain the cartesian equation.

From the parametric equation $x = a \tan t$, we have $\tan t = \frac{x}{a}$.
Substitute this into the expression 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, $\sqrt{a^2} = a$. Therefore:
$y = \frac{x}{a}\sqrt{a^2 + x^2}$

1. Prove that the function is odd by evaluating f(-x) using the Cartesian equation from part (b).

From part (b), the cartesian equation is $\mathrm{f}(x) = x\sqrt{1 + \left(\frac{x}{a}\right)^2}$.
$\mathrm{f}(-x) = (-x)\sqrt{1 + \left(\frac{-x}{a}\right)^2}$
$\mathrm{f}(-x) = -x\sqrt{1 + \frac{x^2}{a^2}}$
$\mathrm{f}(-x) = -\mathrm{f}(x)$

2. Determine the domain and range boundaries of the sketch from the given parametric parameter $t$.

Given $-\frac{\pi}{4} < t < \frac{\pi}{4}$ and $x = a \tan t$:
$a \tan\left(-\frac{\pi}{4}\right) < x < a \tan\left(\frac{\pi}{4}\right)$
$-a < x < a$
When $x \to \pm a$, $y \to \pm a\sqrt{1 + 1} = \pm a\sqrt{2}$.
These boundary points are not included in the curve.

3. Sketch the curve, correctly reflecting the odd symmetry, strict domain, and general shape.

Draw axes indicating the origin $(0,0)$.
Mark the open endpoints at $(-a, -a\sqrt{2})$ and $(a, a\sqrt{2})$.
Sketch an increasing curve passing through the origin. Ensure the portion in the first quadrant is rotationally symmetric to the portion in the third quadrant (concave up for $x>0$, concave down for $x<0$).

1. Derive the cartesian equation for $C$ from part (b) as it is required for the integration.

From the parametric equations, $x = a \tan t \implies \tan t = \frac{x}{a}$.
Given $y = a \sec^2 t \sin t$, we substitute $\sec^2 t = 1 + \tan^2 t$ and use the result from (a), $\sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}}$:
$y = a (1 + \tan^2 t) \left( \frac{\tan t}{\sqrt{1 + \tan^2 t}} \right) = a \tan t \sqrt{1 + \tan^2 t}$
Substituting $\tan t = \frac{x}{a}$:
$\mathrm{f}(x) = a \left(\frac{x}{a}\right) \sqrt{1 + \left(\frac{x}{a}\right)^2} = x \sqrt{\frac{a^2 + x^2}{a^2}} = \frac{x}{a} \sqrt{a^2 + x^2}$

2. Set up the definite integral for the specified area, applying the symmetry property from part (c).

Using the result from (c) that $\mathrm{f}(-x) = -\mathrm{f}(x)$, the curve has rotational symmetry about the origin. The region bounded between $x = -A$ and $x = 0$ lies below the $x$-axis, but its area is identical in size to the region between $x = 0$ and $x = A$.
The total exact area bounded by $C$ and the $x$-axis is therefore twice the area from $x = 0$ to $x = A$:
$\text{Area} = 2 \int_0^A \mathrm{f}(x) \, dx = 2 \int_0^A \frac{x}{a} \sqrt{a^2 + x^2} \, dx$

3. Integrate and evaluate the limits to find the exact area in terms of $A$ and $a$.

Evaluate the integral using the reverse chain rule (or substitution), recognizing that $\frac{d}{dx} (a^2 + x^2)^{3/2} = 3x (a^2 + x^2)^{1/2}$:
$\text{Area} = \frac{2}{a} \left[ \frac{1}{3} (a^2 + x^2)^{3/2} \right]_0^A$
Substitute the upper limit $A$ and lower limit $0$:
$= \frac{2}{3a} \left( (a^2 + A^2)^{3/2} - (a^2 + 0)^{3/2} \right)$
Since $a$ is a positive constant, $(a^2)^{3/2} = a^3$, yielding:
$= \frac{2}{3a} \left( (a^2 + A^2)^{3/2} - a^3 \right)$