= Solution
In the <center-of-momentum frame>, <four-momentum conservation> gives two opposite <photon> momenta, each of <energy> $M_\pi/2$. Let $z=\cos\theta_*$, with the axis along the <Lorentz boost> from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The <Lorentz transformation> of <four-momentum> gives
$$
E_1=\frac{\gamma M_\pi}{2}(1+\beta z),\qquad
E_2=\frac{\gamma M_\pi}{2}(1-\beta z).
$$
Writing $s=\sin(A/2)$, <2/b> now implies
$$
s^2=\frac1{\gamma^2(1-\beta^2z^2)},\qquad
|z|=\frac{\sqrt{\gamma^2s^2-1}}{\beta\gamma s}.
$$
The <isotropy> assumption is $dN=a\,|dz|$ on $-1\le z\le1$. Each interior opening angle has \b[two] inverse branches, $z$ and $-z$, both of which must enter the <change of variables for probability densities>. The positive branch has <derivative>
$$
\left|\frac{dz}{dA}\right|
=\frac{\cos(A/2)}{2\beta\gamma\sin^2(A/2)\sqrt{\gamma^2\sin^2(A/2)-1}}.
$$
Consequently the <diphoton opening-angle distribution> is
$$
\boxed{\frac{dN}{dA}=\frac a\gamma\frac{\cos(A/2)}{\beta\sin^2(A/2)\sqrt{\gamma^2\sin^2(A/2)-1}},\qquad f(\gamma)=\gamma^{-1}.}
$$
Its <support> is $2\arcsin(1/\gamma)<A<\pi$ for $0<\beta<1$, and its integral is $2a$, the integral of the original angular <probability density> before normalization. Thus the normalized <probability density> is $(2a)^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.
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