4. Given in $\triangle A B C$, $\angle A=120^{\circ}$, let vector $\alpha=\frac{\overrightarrow{B A}}{|\overrightarrow{B A}| \cos A}+\frac{\overrightarrow{B C}}{|\overrightarrow{B C}| \cos C}, \beta=$ $\frac{\overrightarrow{C A}}{|\overrightarrow{C A}| \cos A}+\frac{\overrightarrow{C B}}{|\overrightarrow{C B}| \cos B}$, then the angle between $\alpha$ and $\beta$ is -. $\qquad$
4. $60^{\circ}$.

Notice that $\boldsymbol{\alpha} \cdot \overrightarrow{C A}=|\overrightarrow{C A}|-|\overrightarrow{C A}|=0, \boldsymbol{\beta} \cdot \overrightarrow{B A}=|\overrightarrow{B A}|-|\overrightarrow{B A}|=0$. That is, $\boldsymbol{\alpha} \perp \overrightarrow{C A}, \boldsymbol{\beta} \perp \overrightarrow{B A}$, thus the angle between $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ should be equal to or supplementary to $\angle A$, i.e., $60^{\circ}$ or $120^{\circ}$.
Also,
$$
\begin{aligned}
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}= & \frac{\overrightarrow{B A} \cdot \overrightarrow{C A}}{|\overrightarrow{B A}| \cdot|\overrightarrow{C A}| \cos ^{2} A}+\frac{\overrightarrow{B A} \cdot \overrightarrow{C B}}{|\overrightarrow{B A}| \cdot|\overrightarrow{C B}| \cos A \cos B} \\
& +\frac{\overrightarrow{B C} \cdot \overrightarrow{C A}}{|\overrightarrow{B C}| \cdot|\overrightarrow{C A}| \cos C \cos A}+\frac{\overrightarrow{B C} \cdot \overrightarrow{C B}}{|\overrightarrow{B C}| \cdot|\overrightarrow{C B}| \cos B \cos C} \\
= & \frac{1}{\cos B \cos C}-\frac{1}{\cos A},
\end{aligned}
$$

Clearly, $\cos B \cos C>0>\cos A$, then $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}>0$, therefore, the angle between $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ equals $60^{\circ}$.