Concurrency of normals If $\theta_1$, $\theta_2$, and $\theta_3$ are the eccentric angles of three points on the hyperbola $x^2/a^2 - y^2/b^2 = 1$ such that
$\sin (\theta_1+\theta_2) +
 \sin (\theta_2+\theta_3) +
 \sin (\theta_3+\theta_1) = 0$ then prove that the normals at these points are concurrent.
P.S. My attempt was to use the same idea as for an ellipse here, but then I realised that it wouldn't do for quite obvious reasons.
Edit 1: As a clarification, do NOT use the hyperbola $xy = c^2$ to answer this question. You may use ONLY the hyperbola I mentioned above.
Edit 2: Here are my efforts so far....
Assume that the equation of normal in parametric form is
$$\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$$
Then, taking $\tan \frac{\theta}{2} = t$ and using the familiar trigonometric identities gives a biquadratic equation in $t$ as:
$$ bkt^4 + 2(a^2 + b^2 + ah)t^3 + 2(a^2 + b^2 - ah)t - bk = 0$$
where we have substituted $(h,k)$ for $(x,y)$ as the point of intersection of the four normals at $\theta_i$ ($i= 1,2,3,4$) as the roots of the above equation.
Observe that this gives
$$\theta_1 + \theta_2 + \theta_3 + \theta_4 = 2n\pi, n \in \mathbb Z$$
but it doesn't get me near the condition at all. I cannot 'eliminate' one root to get a cubic having three roots. I need some help, particularly in eliminating one root. Any efforts towards this could be welcome.
Other attempts I tried include complex numbers and using the fact that if we have a repeated root, then the equation will have its slope zero at this particular root.
Edit 3: In reply to Blue's comment, the point of intersection comes out to be
$$P \equiv
\left(
  \frac{(a^2 + b^2)\sec \theta_1 \sec \theta_2 \cos
        \left(
          \frac{\theta_1-\theta_2}{2}
        \right)}
       {a\cos
        \left(
          \frac{\theta_1 + \theta_2}{2}
        \right)},
  -\frac{(a^2 + b^2)\tan \theta_1 \tan \theta_2 \tan
         \left(
           \frac{\theta_1 + \theta_2}{2}
         \right)}{b}
\right)$$
On substituting in the third equation, we get
$$
\sin^2 \theta_1 \cos^2 \theta_2 \sin \theta_3 \cos \theta_3-
\cos^2 \theta_1 \sin^2 \theta_2 \sin \theta_3 \cos \theta_3+
\\ \quad
\sin^2 \theta_1 \sin \theta_2 \cos \theta_2 \sin \theta_3 \cos \theta_3+
\sin \theta_1 \cos^2 \theta_1 \sin ^2 \theta_2 \sin \theta_3-
\\ \quad
\sin^2 \theta_1 \sin \theta_2 \cos^2 \theta_2 \cos \theta_3-
\sin \theta_1 \cos \theta_1 \sin^2 \theta_2 \cos \theta_2 \cos \theta_3
\\ =
\sin^2 \theta_1 \cos \theta_1 \cos^2 \theta_2 \cos \theta_3+
\sin \theta_1 \cos^2 \theta_1 \sin \theta_2 \cos \theta_3 \cos \theta_3-
\\ \quad
\sin \theta_1 \cos \theta_1 \sin \theta_2 \cos^2 \theta_2 \cos \theta_3-
\cos^2 \theta_1 \sin^2 \theta_2 \cos \theta_2 \cos \theta_3$$
Now I am unable to simplify. Please note that this has been homogenised by using the fact that $\sin^2 \theta + \cos^2 \theta = 1$. Yet I am unable to get to the hypothesis.
 A: Start with $$\frac{ax}{\text{sec}\ \theta_1} + \frac{by}{\text{tan}\ \theta_1} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_2} + \frac{by}{\text{tan}\ \theta_2} = a^2 + b^2$$
$$\frac{ax}{\text{sec}\ \theta_3} + \frac{by}{\text{tan}\ \theta_3} = a^2 + b^2$$
When are these three straight lines concurrent?
Equate the following determinant to $0$ and simplify!
\begin{vmatrix}
\frac{a}{\sec \theta_1} & \frac{b}{\tan \theta_1} & a^2 + b^2  \\ 
\frac{a}{\sec \theta_2} & \frac{b}{\tan \theta_2} & a^2 + b^2  \\ 
\frac{a}{\sec \theta_3} & \frac{b}{\tan \theta_3} & a^2 + b^2   \notag
\end{vmatrix}
Note:
In other words, you need to equate the following to $0$ and simplify:
\begin{vmatrix}
\frac{1}{\sec \theta_1} & \frac{1}{\tan \theta_1} & 1  \\ 
\frac{1}{\sec \theta_2} & \frac{1}{\tan \theta_2} & 1  \\ 
\frac{1}{\sec \theta_3} & \frac{1}{\tan \theta_3} & 1   \notag
\end{vmatrix}
A few more steps:
$1.$ Multiply $R_1$ by $\sin \theta_1$, $R_2$ by $\sin \theta_2$ and $R_3$ by $\sin \theta_3$
$2.$ Try to show that the above simplifies to 
$-2 \sin \frac{\theta_1 - \theta_2}{2} \sin \frac{\theta_2 - \theta_3}{2} \sin \frac{\theta_3 - \theta_1}{2}[\sin (\theta_1 + \theta_2) + \sin (\theta_2 + \theta_3) + \sin (\theta_3 + \theta_1)] = 0$
$3.$ Argue that $\sin (\theta_1 + \theta_2) + \sin (\theta_2 + \theta_3) + \sin (\theta_3 + \theta_1) = 0$
In fact, you need to prove the converse of this result. That is also true and you can write the proof "bottom-up"
