= Solution
For $u>0$, take the initial electron direction as the longitudinal axis and put $\beta=u/c$. Incoming <four-momentum conservation> gives
$$
h(\nu_1+\nu_2)=mc^2(\gamma+1),\qquad
\frac hc(\nu_1e_1+\nu_2e_2)=m\gamma u\,\hat u.
$$
The transverse components must cancel. In the noncollinear case the two transverse directions are opposite, and $\nu_1\sin\theta_1=\nu_2\sin\theta_2$. Hence the ratio of longitudinal momentum times $c$ to energy is
$$
\frac{\gamma\beta}{\gamma+1}
=\frac{\nu_1\cos\theta_1+\nu_2\cos\theta_2}{\nu_1+\nu_2}
=\frac{\sin(\theta_1+\theta_2)}{\sin\theta_1+\sin\theta_2}.
$$
Writing $s=(\theta_1+\theta_2)/2$ and $d=(\theta_1-\theta_2)/2$, the last ratio is $\cos s/\cos d$. It also equals $(1+\cos(\theta_1+\theta_2))/(\cos\theta_1+\cos\theta_2)$ whenever that denominator is nonzero. Since $\gamma^2\beta^2=\gamma^2-1$, the <angular relation for two-photon electron-positron annihilation> is
$$
\boxed{\frac{1+\cos(\theta_1+\theta_2)}{\cos\theta_1+\cos\theta_2}
=\frac{\gamma\beta}{\gamma+1}
=\sqrt{\frac{\gamma-1}{\gamma+1}}.}
$$
The positive root follows from $u>0$. For noncollinear emission the positive momentum-to-energy ratio ensures $\cos s>0$, and $|d|<\pi/2$ ensures the stated denominator is positive.
The printed quotient needs a qualification for exactly collinear <photons>. The allowed configuration $\theta_1=0$, $\theta_2=\pi$ has positive frequencies
$$
h\nu_{1,2}=\frac{mc^2}{2}\left[(\gamma+1)\pm\gamma\beta\right],
$$
which obey both conservation equations, but the printed left side is $0/0$. Thus the quotient holds for noncollinear angles, or as its appropriate noncollinear limit. The undivided identity
$$
1+\cos(\theta_1+\theta_2)
=\sqrt{\frac{\gamma-1}{\gamma+1}}(\cos\theta_1+\cos\theta_2)
$$
remains true in the collinear case, where both sides vanish. The limit $u=0$ similarly requires care because the initial direction is then undefined and the <photons> are antiparallel.
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