The second root, find the locus of point $(a, b)$.

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Solve according to Vieta's theorem
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sec \theta+\csc \theta=a, \\
\sec \theta \cdot \csc \theta=b
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
\sin \theta \cdot \cos \theta=\frac{1}{b}, \\
\sin \theta+\cos \theta=\frac{a}{b}
\end{array}\right.
\end{array}
$$

Eliminating $\theta$, we have $\frac{a^{2}}{b^{2}}=1+\frac{2}{b}$,
which is $(b+1)^{2}-a^{2}=1$. However, the trajectory of $(a, b)$ is not the entire hyperbola.
$$
\begin{array}{c}
\because \quad \sin 2 \theta=\frac{2}{b},-1 \leqslant \frac{2}{i} \leqslant 1 . \\
\therefore \quad U \geqslant 2 \text { though } b \leqslant-2
\end{array}
$$

If we take $a$ as the x-axis and $b$ as the y-axis, then the constraints obtained indicate that the required trajectory is the lower half of the hyperbola and the upper half where $b \geqslant 2$.

The restrictions caused by the range of trigonometric function values are very common and should be given sufficient attention in teaching.