Relativistic particle annihilation- getting wrong answer While preparing for my exams, I found the following question on a past paper for which I am getting a different answer to what the question says I should be getting, I can't see where I am going wrong or whether what I have is equivalent to the answer but I suspect it isn't. (Note: This is not for an assignment, it is just some extra revision I am doing to be better prepared).
Question: 
 In the laboratory frame an electron travelling with velocity $u$ collides with a positron at rest. They annihilate, producing two photons of frequencies $ν_1$ and $ν_2$ that
move off at angles $θ_1$ and $θ_2$ to $u$, in the directions of the unit vectors $e_1$ and $e_2$ respectively.
By considering 4-momenta in the laboratory frame, or otherwise, show that $\dfrac{1+\cos(\theta_1+\theta_2)}{\cos(\theta_1)+\cos(\theta_2)} = \sqrt{\dfrac{\gamma-1}{\gamma+1}}$ where $\gamma$ is the Lorentz factor for the electron.
Here is what I have achieved so far:
Let $m$ be the mass of the positron/election. Consider the frame in which the election is initially travelling in the x-axis. By conservation of 4-momentum: $$ m\gamma\pmatrix{c \\ u \\ 0 \\ 0} + m\pmatrix{c \\ 0 \\ 0 \\ 0} = \frac{h\nu_1}{c}\pmatrix{1 \\ \cos(\theta_1) \\ \sin(\theta_1) \\ 0} + \frac{h\nu_2}{c}\pmatrix{1 \\ \cos(\theta_2) \\ \sin(\theta_2) \\ 0}$$
The ratio of the second and first lines gives us: $$ \frac{\gamma u}{(1+\gamma)c} = \frac{\nu_1\cos(\theta_1)+\nu_2\cos(\theta_2)}{\nu_1+\nu_2}$$
But $  (\frac{\gamma u}{(1+\gamma)c})^2 = \frac{1}{(1+\gamma)^2}(\frac{\gamma u}{c})^2$ and $(\frac{\gamma u}{c})^2 = \gamma^2 u^2/c^2$ = $\frac{u^2/c^2}{1-u^2/c^2} = -1 + \frac{1}{1-u^2/c^2} = \gamma^2-1$. So $(\frac{\gamma u}{(1+\gamma)c})^2 = \frac{\gamma^2-1}{(\gamma+1)^2} = \frac{\gamma-1}{\gamma+1}$. So $\frac{\gamma u}{(1+\gamma)c} = \sqrt{\frac{\gamma-1}{\gamma+1}}$. So we have:
$$ \sqrt{\frac{\gamma-1}{\gamma+1}} = \frac{\nu_1\cos(\theta_1)+\nu_2\cos(\theta_2)}{\nu_1+\nu_2}$$
Now the third line in the Conservation of 4-momentum equation tells us: $\frac{h}{c}(\nu_1\sin(\theta_1)+\nu_2\sin(\theta_2)) = 0$ So $\nu_2 = -\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}$ (division by $0$ not a problem here as that would imply the frequency of one of the photons was $0$).
Putting this in to our result from before (and cancelling $\nu_1$ and simplifying): 
$$ \sqrt{\frac{\gamma-1}{\gamma+1}} = \frac{\nu_1\cos(\theta_1)-\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}\cos(\theta_2)}{\nu_1-\nu_1\frac{\sin(\theta_1)}{\sin(\theta_2)}} = \frac{\cos(\theta_1)-\frac{\sin(\theta_1)}{\sin(\theta_2)}\cos(\theta_2)}{1-\frac{\sin(\theta_1)}{\sin(\theta_2)}} = \frac{\cos(\theta_1)\sin(\theta_2)-\sin(\theta_1)\cos(\theta_2)}{\sin(\theta_2)-\sin(\theta_1)} = \frac{\sin(\theta_2-\theta_1)}{\sin(\theta_2)-\sin(\theta_1)}$$
However this is not the result we are supposed to prove. Can anyone see where I went wrong with this?
Thanks in advance.
 A: If you need a minus sign, put it.
$$m\gamma\pmatrix{c \\ u \\ 0 \\ 0} + m\pmatrix{c \\ 0 \\ 0 \\ 0} = \frac{h\nu_1}{c}\pmatrix{1 \\ \cos(\theta_1) \\ \sin(\theta_1) \\ 0} + \frac{h\nu_2}{c}\pmatrix{1 \\ \cos(\theta_2) \\ -\sin(\theta_2) \\ 0}$$
You suppose the angles are defined both the same way, anticlockwise from the $x$ axis, but usually in collision exercises, one is measured clockwise and the other anticlockwise (e.g.)
The $u=0$ thing is not a problem as you take limits ($u\to0$), the sum can be  finite and the numerator is clearly, with the "redefinition", zero. And the point about the possibility the formula gives for the rhs to be greater than $1$, you almost said it: some angles are forbidden. We need more information for a complete/determinate description or this is impossible, at most this relation we are commenting between them (not sure about possible/impossible).
I finally succeed proving the hint Jens gave, but it took me a too long while. I think there is a simpler way to prove it.
