Prove that by taking the bisector of the angle formed by rays $l_{1}$ and $l_{2}$ as the $x$-axis, and the line passing through point $O$ and perpendicular to the $x$-axis as the $y$-axis, we establish a Cartesian coordinate system.
Let the moving point $X\left(x_{0}, y_{0}\right)$,
ray $l_{1}: y=k x(k>0)$,
$$\begin{array}{l} l_{2}: y=-k x(k>0), \\ |O A|=\rho_{1},|O B|=\rho_{2} . \end{array}$$

Then, from $S_{\triangle O A X}=S_{\triangle O B X}=\frac{c}{2}$, we get
$$\begin{array}{l} \frac{1}{2} \cdot \frac{\left|k x_{0}-y_{0}\right|}{\sqrt{1+k^{2}}} \cdot \rho_{1}=\frac{c}{2}, \\ \frac{1}{2} \cdot \frac{\left|-k x_{0}-y_{0}\right|}{\sqrt{1+k^{2}}} \cdot \rho_{2}=\frac{c}{2} . \\ \text { Thus, } \rho_{1}=\frac{c \sqrt{1+k^{2}}}{\left|k x_{0}-y_{0}\right|}, \rho_{2}=\frac{c \sqrt{1+k^{2}}}{\left|k x_{0}+y_{0}\right|} . \\ \text { Also, } \frac{1}{2} \rho_{1} \rho_{2} \sin 2 \theta=c \text {, then } \\ \frac{1}{2} \cdot \frac{c^{2}\left(1+k^{2}\right)}{\left|k^{2} x_{0}^{2}-y_{0}^{2}\right|} \cdot \frac{2 k}{1+k^{2}}=c . \end{array}$$

Since point $X$ is within the region enclosed by the rays, we have
$$\begin{array}{l} \frac{1}{2} \cdot \frac{c^{2}\left(1+k^{2}\right)}{k^{2} x_{0}^{2}-y_{0}^{2}} \cdot \frac{2 k}{1+k^{2}}=c \\ \Rightarrow k^{2} x_{0}^{2}-y_{0}^{2}=c k\left(x_{0}>0\right) \end{array}$$

which is the equation of the trajectory $\Gamma$.
From the equation, it is easy to see that the conclusion holds.