Answer
Expression of $\tan\theta$ in terms of $\tan\alpha$ and $\tan\beta$.
$$\tan\theta=\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$$
Work Step by Step
As in the previous exercise, we deduce that $$\theta=\beta-\alpha$$
which means $$\tan\theta=\tan(\beta-\alpha)$$
Apply the identity of tangent difference:
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Therefore, $$\tan\theta=\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$$
This is the expression of $\tan\theta$ in terms of $\tan\alpha$ and $\tan\beta$.