May I assume that an undefined fraction has a denominator equal to 0?

Question

It is given that $ \tan \alpha = \frac{ \tan \theta_2 - \tan \theta_1 }{ 1 + \tan \theta_2 \tan \theta_1 } $.

If $ \alpha = \frac{\pi}{2} $, meaning that $ \tan \alpha $ is undefined, is the RHS also undefined? If so, could I then deduce that $ 1 + \tan \theta_2 \tan \theta_1 = 0$?

Answer 1

The formula on the right hand side is of $\color{red}{\tan(\theta_2-\theta_1)}$.

So, we have $\color{blue}{\theta_2-\theta_1 = \alpha}$

If $\alpha = \frac{\pi}{2}$, then $\theta_2 - \theta_1 = \frac{\pi}{2} \implies \color{blue}{\theta_1 = \theta_2-\frac\pi2}$

$$\tan\theta_1 = \tan(\theta_2-\frac\pi2) = -\cot\theta_2 = -\frac1{\tan\theta_2} \implies\tan\theta_1\tan\theta_2 = - 1$$

$$1+\tan\theta_1\tan\theta_2 = 0$$