Showing $\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}}$

Let $x$ be any of $\alpha_1$, $\beta_1$, or $\gamma_1$ and suppose $\alpha_1+\beta_1+\gamma_1=\pi$. Then

$$e^{i(\alpha_2+\beta_2)}+e^{i(\alpha_2+\gamma_2)}+e^{i(\beta_2+\gamma_2)}=1\qquad \left(x \in \mathbb{R} : \frac{x + \pi}{2\pi} \not\in \mathbb{Z}\right)\tag{1}$$

or

$$\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}},\qquad \left(x = 2\pi n + \pi, \quad n \in \mathbb{Z}\right) \tag{2}$$

where complex $\alpha_2=\cos^{-1}(\csc(\alpha_1))$ and similarly for $\beta_2$ and $\gamma_2$.

The identities $(1-2)$ are consequences of Theorem 1 in this blog post and the trigonometric identities 3 and 4 (counting from top to bottom) in this list.

Addendum: generalize $(1-2)$.