The normal to the curve $y = 2\tan 3x - 1$ at point $P\left(\frac{\pi}{3}, -1\right)$ crosses the $y$-axis at point $A$ and the $x$-axis at $B$. Find the exact area of triangle $AOB$.

1. Differentiate the equation of the curve to find the gradient function.

$y = 2\tan 3x - 1$
$\frac{dy}{dx} = 2(3\sec^2 3x) = 6\sec^2 3x$

2. Evaluate the gradient of the tangent at $x = \frac{\pi}{3}$.

At $x = \frac{\pi}{3}$,
$\frac{dy}{dx} = 6\sec^2\left(3 \times \frac{\pi}{3}\right) = 6\sec^2(\pi)$
Since $\cos(\pi) = -1$, $\sec(\pi) = -1$.
$\frac{dy}{dx} = 6(-1)^2 = 6$
So the gradient of the tangent is $m_t = 6$.

3. Find the gradient of the normal.

Gradient of normal $m_n = -\frac{1}{m_t} = -\frac{1}{6}$.

4. Find the equation of the normal at $P\left(\frac{\pi}{3}, -1\right)$.

$y - (-1) = -\frac{1}{6}\left(x - \frac{\pi}{3}\right)$
$y + 1 = -\frac{1}{6}x + \frac{\pi}{18}$
$y = -\frac{1}{6}x + \frac{\pi}{18} - 1$

5. Find the coordinates of $A$ ($y$-intercept) and $B$ ($x$-intercept).

At $A$, $x = 0 \implies y = \frac{\pi}{18} - 1$. So $A\left(0, \frac{\pi}{18} - 1\right)$.
At $B$, $y = 0 \implies 0 = -\frac{1}{6}x + \frac{\pi}{18} - 1$
$\frac{1}{6}x = \frac{\pi}{18} - 1$
$x = \frac{\pi}{3} - 6$. So $B\left(\frac{\pi}{3} - 6, 0\right)$.

6. Calculate the exact area of triangle $AOB$.

Area $= \frac{1}{2} \times |x_B| \times |y_A|$
Since $\pi \approx 3.14$, $\frac{\pi}{18} - 1 < 0$ and $\frac{\pi}{3} - 6 < 0$.
$|y_A| = 1 - \frac{\pi}{18}$ and $|x_B| = 6 - \frac{\pi}{3}$.
Area $= \frac{1}{2} \left(6 - \frac{\pi}{3}\right) \left(1 - \frac{\pi}{18}\right)$
Notice that $1 - \frac{\pi}{18} = \frac{1}{6}\left(6 - \frac{\pi}{3}\right)$.
Area $= \frac{1}{2} \left(6 - \frac{\pi}{3}\right) \times \frac{1}{6}\left(6 - \frac{\pi}{3}\right) = \frac{1}{12}\left(6 - \frac{\pi}{3}\right)^2$
Alternatively, $\frac{1}{12}\left(\frac{\pi}{3} - 6\right)^2$.