Cutting the quadrilateral along an appropriate diagonal (say, of length $p$) gives triangles with sides $a$, $d$, $p$ and $b$, $c$, $p$. By the Spherical Law of Cosines, we can write
$$\cos a \cos d + \sin a \sin d \cos \alpha = \cos p = \cos b \cos c+\sin b \sin c \cos \gamma \tag1$$
Or, using the shorthand $\overline{x}:=\sin x$, $\ddot{x}:=\cos x$,
$$\ddot{a} \ddot{d} + \overline{a} \overline{d} \ddot{\alpha} = \ddot{b} \ddot{c}+\overline{b}\overline{c}\ddot{\gamma}$$
Thus, defining $x_2:=\frac12x$ for convenience, we can expand $\cos\alpha$ and $\cos\gamma$ using two versions of the half-angle identities $\cos\theta=1-2\sin^2\theta_2=2\cos^2\theta_2-1$:
$$\begin{align}
\ddot{a} \ddot{d} + \overline{a} \overline{d} \;(1-2\overline{\alpha_2}^2) &= \ddot{b} \ddot{c} + \overline{b} \overline{c}\;(2\ddot{\gamma_2}^2-1) \tag2 \\[4pt]
\left(\ddot{a} \ddot{d} + \overline{a} \overline{d}\right) - \left(\ddot{b} \ddot{c}-\overline{b} \overline{c}\right) &=
2\overline{a}\overline{d}\overline{\alpha_2}^2+2\overline{b}\overline{ c}\ddot{\gamma_2}^2
\tag3\\[4pt]
\cos(a-d)- \cos(b+c) &=
2\overline{a}\overline{d}\overline{\alpha_2}^2+2\overline{b}\overline{ c}\ddot{\gamma_2}^2
\tag4\\[4pt]
2 \sin(-a_2 +b_2 + c_2 + d_2) \sin(a_2 + b_2 + c_2 -d_2) &=
2\overline{a}\overline{d}\overline{\alpha_2}^2+2\overline{b}\overline{ c}\ddot{\gamma_2}^2
\tag5
\end{align}$$
Thus, with $s:=\frac12(a+b+c+d)$,
$$\sin a\sin d\sin^2\alpha_2+\sin b\sin c\cos^2\gamma_2
\;=\; \sin(s-a) \sin(s-d) \tag{$\star$}$$
Likewise,
$$\sin a\sin d \cos^2\alpha_2 + \sin b \sin c \sin^2\gamma_2 \;=\; \sin(s-b)\sin(s-c) \tag{$\star\star$}$$
FYI, in hyperbolic geometry, the counterpart relations simply replace "$\sin(\text{length})$" with "$\sinh(\text{length})$".
Investigation of consequent "classical metric relations" is left as an exercise to the reader.
