Here is what I found after some investigation.
My ghastly expression is $$\begin{align}v_{\text{0A}}^2=\frac{2 \sin (\alpha +\beta )}{2 \sin (\alpha ) \sin (\alpha +\beta )-\frac{\cos (\beta ) \left(\sin ^2(\beta )-\sin ^2(\alpha )\right)}{\sin ^2(\beta )}}.\end{align}$$ Your (neater) expression is $$\begin{align} v_{\text{0B}}^2=\frac{4 \sin ^2(\beta )}{3 \sin (\alpha )-\sin (\alpha +2 \beta )}.\end{align}$$ The expression in choice A with the denominator multiplied by -1 to make it positive when ##\alpha=\beta~## is $$\begin{align}v_{\text{0C}}^2=\frac{\cos (\alpha ) \left(2 \tan ^2(\alpha ) \tan ^2(\beta ) (\tan (\alpha )+\tan (\beta ))\right)}{\sin ^2(\alpha ) \left(2 \tan (\alpha ) \tan ^3(\beta )+2 \tan ^2(\alpha ) \tan ^2(\beta )-\tan ^2(\alpha )+\tan ^2(\beta )\right)}.\end{align}$$
I. Equations (1) and (2) give identical results but equation (3) does not. I verified this by fixing ##\alpha## to some value and plotting ##v0^2## as a function of ##\beta## from ##0## to ##\pi/2##. The graphs for (1) and (2) were on top of each other whilst the graph for (3) was not. From now on I will refer to equations (1) and (2) collectively as "our equation" and equation (3) as "their equation" by replacing sines and cosines with tangents.
II. The condition for our equation to be positive is $$\tan(\alpha)>\frac{\sin(\beta)\cos(\beta)}{1+\sin^2(\beta)}.$$The ratio on the right has a maximum value of 0.3535 which translates to ##\alpha > 19.5^o.## In other words, as long as the above condition is met, our equation will always have a solution. This is not true for their equation which changes sign in the region ##0<\beta<\pi/2## no matter what one chooses for ##\alpha##. I verified this last result by doing several plots. The one below is for ##\alpha = 21.6^o##. The blue line is our equation and the orange line is their equation. The vertical line marks the value of ##\alpha##. It is ironic and instructive to see that our equation matches theirs only at ##\alpha=\beta## and ##\beta=\frac{\pi}{2}##, i.e. both equations pass the tests that I proposed in post #12.
III. Both our and their equations agree when ##\alpha = \beta## and when ##\beta = \frac{\pi}{2}.## One can get around the "undefined" problem with their equation (3) if one divides both numerator and denominator by ##\tan^3(\beta)## and then take the limit as ##\tan(\beta)\rightarrow \infty.##
Edits: After some algebra, I was able to transform equation (1) into equation (2) so they
are identical. I wonder if it is worth anyone's while to see how close one can get to their equation.
Also edited to add the vertical line at ##\beta =\alpha## in the plot.
Final edit: Equations (1) and (2) can be algebraically transformed to one of the options with the negative signs in front of the factors of ##2## in the denominator flipped to positive signs.