This identity stems from:$$\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha \tan\beta \tan\gamma$$
(for $\alpha + \beta + \gamma = \pi$) which can easily be proven using the $\tan(\alpha + \beta)$ identity.
However, I have been unable to prove $$\sin\alpha \cos\alpha + \sin\beta \cos\beta + \sin\gamma \cos\gamma = 2\sin\alpha \sin\beta \sin\gamma,$$ where $\alpha + \beta + \gamma = \pi$.
EDIT: I proved it.
From $\alpha + \beta + \gamma = \pi$ we have$\alpha = \pi - (\beta + \gamma)$
This implies
$$
\sin\alpha = \sin(\beta +\gamma) \quad \text{and }\quad \cos\alpha = -\cos(\beta +\gamma).$$
Then, we have
$$\begin{align}
\sin\alpha \cos\alpha &=-\sin(\beta +\gamma)\cos(\beta + \gamma)\\
&= -(\sin\beta \cos\gamma + \sin\gamma \cos\beta)(\cos\beta \cos\gamma - \sin\beta \sin\gamma)\\
&= -(\sin\beta \cos\beta \cos^2\gamma \ - \sin\gamma \cos\gamma \sin^2\beta + \sin\gamma \cos\gamma \cos^2\beta - \sin\beta \cos\beta \sin^2\gamma)\\
&= -[\sin\beta \cos\beta \ (\cos^2\gamma - \sin^2\gamma) + \sin\gamma \cos\gamma \ (\cos^2\beta - \sin^2\beta)]\\
&= \sin\beta \cos\beta \ (\sin^2\gamma - \cos^2\gamma) + \sin\gamma \cos\gamma \ (\sin^2\beta\ - \cos^2\beta )
\end{align}$$
Substituting this in $\sin\alpha \cos\alpha + \sin\beta \cos\beta + \sin\gamma \cos\gamma$ we get
$$\begin{align}
\sin\beta \cos\beta &(\sin^2\gamma - \cos^2\gamma) + \sin\beta \ cos\beta + \sin\gamma \cos\gamma \ (\sin^2\beta\ - cos^2\beta) + \sin\gamma \cos\gamma \\
&= \sin\beta \cos\beta \ (\sin^2\gamma +1 - \cos^2\gamma) + \sin\gamma \cos\gamma \ (\sin^2\beta +1 - \cos^2\beta)\\
&= \sin\beta \cos\beta \ (\sin^2\gamma + \sin^2\gamma) + \sin\gamma \cos\gamma \ (\sin^2\beta +\sin^2\beta)\\
&= \sin\beta \cos\beta \ (2\sin^2\gamma) + \sin\gamma \cos\gamma \ (2\sin^2\beta)\\
&= 2\sin\beta \sin\gamma \ (\sin\gamma \cos\beta + \sin\beta \cos\gamma)\\
&= 2\sin\beta \sin\gamma \sin(\beta + \gamma)\\
&= 2\sin\beta \sin\gamma \sin\alpha\\
&= 2\sin\alpha \sin\beta \sin\gamma.
\end{align}$$
