Answer
$$\tan\theta=\frac{m_2-m_1}{1+m_1m_2}$$
Work Step by Step
From the previous exercise, it has been found that
$$\tan\theta=\frac{\tan\beta-\tan\alpha}{1+\tan\alpha\tan\beta}$$
Now we have $m_1$ is the slope of the line that makes an angle of $\alpha$ with $Ox$ and $m_2$ is the slope of the line that makes an angle of $\beta$ with $Ox$.
Therefore we can take $\tan\alpha=m_1$ and $\tan\beta=m_2$ and replace back to $\tan\theta$ formula.
$$\tan\theta=\frac{m_2-m_1}{1+m_1m_2}$$