Where does the definition of $\theta$ in an $RLC$ circuit come from. Why is $\theta = \tan^{-1}\left(\frac{1/(\omega C) -\omega L}{R}\right)$ ?

This $\theta$ represents the phase shift between input and output voltages of the four-terminal network. According to $\hat{v}_{ab}(j\omega) = \hat{H}(j\omega) \hat{v}_{\rm in}(j\omega)$, the relative phase between $\hat{v}_{ab}$ and $\hat{v}_{\rm in}$ is the phase of the transfer function $\hat{H}$. So to find this phase we write $\hat{H}(j \omega)$ in polar form, $\hat{H}(j\omega) = \vert\hat{H}\vert e^{j\theta}$. The phase $\theta$ of a complex number $\hat{C} = A+jB$ written as $\vert\hat{C}\vert e^{j\theta}$ is $\theta=\tan^{-1}(B/A)$: it's inverse tangent of the imaginary part over the real part. For the $RLC$ network, $\hat{H}(j\omega) = \frac{R}{R+j(\omega L - \frac{1}{\omega C})}$. This can be rewritten (multiply top and bottom by $R-j(\omega L - \frac{1}{\omega C})$) as $\hat{H}(j\omega) = \frac{R^2-j(\omega L- \frac{1}{\omega C})R}{R^2+(\omega L - \frac{1}{\omega C})^2}$. Hence, the phase shift for the network is $\theta = \tan^{-1}\left(-(\omega L - \frac{1}{\omega c})R/R^2\right) = \tan^{-1}\left(\frac{\frac{1}{\omega C} - \omega L}{R}\right)$. We'll see lots of examples like this, so work through the algebra if you are uncomfortable with it.