A curve $C$ is defined parametrically by $x = a \tan t, \quad y = a \sec^2 t \sin t, \quad -\frac{\pi}{4} < t < \frac{\pi}{4},$ where $a$ is a positive constant.

1. Apply the Pythagorean identity $1 + \tan^2 t = \sec^2 t$ to the denominator.

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

$= \frac{\tan t}{\sqrt{\sec^2 t}}$

2. Evaluate the square root by considering the domain of $t$.

For $-\frac{\pi}{4} < t < \frac{\pi}{4}$, $\cos t > 0$, so $\sec t > 0$.

Hence, $\sqrt{\sec^2 t} = |\sec t| = \sec t$.

$= \frac{\tan t}{\sec t}$

3. Express the trigonometric functions in terms of sine and cosine to simplify.

$= \frac{\left(\frac{\sin t}{\cos t}\right)}{\left(\frac{1}{\cos t}\right)}$

$= \sin t = \text{RHS} \quad \text{(Shown)}$

1. Express $y$ entirely in terms of $\tan t$ using the given identity from part (a) and the Pythagorean identity for $\sec^2 t$.

From part (a), we have $\sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}}$.
We also know that $\sec^2 t = 1 + \tan^2 t$.
Substituting these into the equation for $y$:
$y = a(1 + \tan^2 t)\left(\frac{\tan t}{\sqrt{1 + \tan^2 t}}\right)$
$y = a \tan t \sqrt{1 + \tan^2 t}$

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

From $x = a \tan t$, we get $\tan t = \frac{x}{a}$.
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 $a > 0$, $\sqrt{a^2} = a$, which yields:
$y = \frac{x}{a}\sqrt{a^2 + x^2}$

1. Show that the function is odd by substituting $-x$ into $\mathrm{f}(x)$ from part (b)

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

2. Determine the endpoints and domain from the parametric restrictions

The parameter $t$ is restricted to $-\frac{\pi}{4} < t < \frac{\pi}{4}$.
As $t \to \frac{\pi}{4}$:
$x \to a \tan\left(\frac{\pi}{4}\right) = a$
$y \to a \sec^2\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) = a(\sqrt{2})^2\left(\frac{1}{\sqrt{2}}\right) = a\sqrt{2}$
As $t \to -\frac{\pi}{4}$, by the odd symmetry proven above, $x \to -a$ and $y \to -a\sqrt{2}$.
Thus, the curve exists for $-a < x < a$ with open endpoints at $(-a, -a\sqrt{2})$ and $(a, a\sqrt{2})$.

3. Describe the sketch of curve $C$

Since $\mathrm{f}(0)=0$ and $\mathrm{f}'(x) > 0$, the curve is strictly increasing and passes through the origin.
The relation $\mathrm{f}(-x) = -\mathrm{f}(x)$ means the graph has rotational symmetry about the origin.
The sketch must show a smooth curve passing through $(0,0)$, starting from an open circle at $(-a, -a\sqrt{2})$ and ending at an open circle at $(a, a\sqrt{2})$.

1. Derive the Cartesian equation of $C$ (finding the required result from part b).

Given $x = a \tan t$, we have $\tan t = \frac{x}{a}$.
From part (a), substitute $\tan t$ into the identity:
$\sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}} = \frac{\frac{x}{a}}{\sqrt{1 + \left(\frac{x}{a}\right)^2}} = \frac{x}{\sqrt{a^2+x^2}}$
Substitute into $y = a \sec^2 t \sin t$:
$y = a(1 + \tan^2 t)\sin t = a\left(1 + \frac{x^2}{a^2}\right)\left(\frac{x}{\sqrt{a^2+x^2}}\right)$
$y = \mathrm{f}(x) = \frac{x}{a}\sqrt{a^2+x^2}$

2. Set up the definite integral for the exact area, using the symmetry established in part (c).

Since $\mathrm{f}(-x) = -\mathrm{f}(x)$ (odd function) and $\mathrm{f}(x) \ge 0$ for $x \ge 0$, the regions below and above the $x$-axis are symmetrical.
The total area is bounded from $x = -A$ 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. Evaluate the integral.

Rewrite the integrand to use the standard integration form $\int \mathrm{f}'(x)[\mathrm{f}(x)]^n \, dx$:
$= \frac{1}{a} \int_{0}^{A} 2x (a^2+x^2)^{\frac{1}{2}} \, dx$
$= \frac{1}{a} \left[ \frac{(a^2+x^2)^{\frac{3}{2}}}{\frac{3}{2}} \right]_0^A$
$= \frac{2}{3a} \left[ (a^2+x^2)^{\frac{3}{2}} \right]_0^A$
$= \frac{2}{3a} \left( (a^2+A^2)^{\frac{3}{2}} - (a^2+0)^{\frac{3}{2}} \right)$
$= \frac{2}{3a} \left( (a^2+A^2)^{\frac{3}{2}} - a^3 \right)$