Addition $\leftrightarrow$ Multiplication

Let $A, B, C$ be real numbers in the range $\left [0, \frac{\pi}{2} \right ) \cup \left (\frac{\pi}{2}, \pi \right]$ such that

$\begin{aligned} \tan A &= x \\ \tan B &= y \\ \tan C &= z. \end{aligned}$

Note that as an angle $\theta$ spans the range $\left [0, \frac{\pi}{2} \right ) \cup \left (\frac{\pi}{2}, \pi \right]$ , $\tan \theta$ spans the range $(-\infty, \infty),$ which is the set of all real numbers, so this substitution is valid.

The given condition rewrites as $\tan A + \tan B + \tan C = \tan A \tan B \tan C.$ We shall prove that this indicates that $A + B + C = k\pi,$ where $k \in \{0, 1, 2, 3\}$ . Manipulating the expression yields

$\begin{aligned} \tan A + \tan B + \tan C &= \tan A \tan B \tan C \\ \tan A + \tan B &= \tan A \tan B \tan C - \tan C \\ \tan A + \tan B &= -\tan C (1 - \tan A \tan B) \\ \dfrac{\tan A + \tan B}{1 - \tan A \tan B} &= -\tan C \\ \tan(A + B) &= -\tan C \\ \tan(A + B) &= \tan(-C). \end{aligned}$

Since the period of tangent is $\pi,$ we can say that $A + B = -C + k\pi$ for some integer $k,$ or $A + B + C = k\pi.$ Because $A, B, C$ are first and second quadrant angles, the possible values of $k$ are 0, 1, 2, and 3.

Now, we look at the second expression. By the tangent double angle identity, we have

$\dfrac{2x}{1 - x^2} = \dfrac{2 \tan A}{1 - \tan^2 A} = \tan 2A.$

Similarly, $\dfrac{2y}{1 - y^2} = \tan 2B$ and $\dfrac{2z}{1 - z^2} = \tan 2C.$

We proved that $A + B + C = k\pi$ previously. Isolating for $C$ yields $C = k\pi - (A + B),$ and multiplying both sides by 2 gives $2C = 2k\pi - (2A + 2B).$ Taking the tangent of both sides, and applying the tangent angle sum identity and the identity $\tan (2k\pi - \theta) = -\tan \theta,$ gives us

$\begin{aligned} \tan 2C &= \tan(2k\pi - (2A + 2B)) \\ \tan 2C &= -\tan(2A + 2B) \\ \tan 2C &= -\dfrac{\tan 2A + \tan 2B}{1 - \tan 2A \tan 2B} \\ \tan 2C(1 - \tan 2A \tan 2B) &= -(\tan 2A + \tan 2B) \\ \tan 2A + \tan 2B + \tan 2C &= \tan 2A \tan 2B \tan 2C \\ \dfrac{2x}{1 - x^2} + \dfrac{2y}{1 - y^2} + \dfrac{2z}{1 - z^2} &= \dfrac{2x}{1 - x^2} \times \dfrac{2y}{1 - y^2} \times \dfrac{2z}{1 - z^2}, \end{aligned}$

as desired. $\blacksquare$