determinant of a symmetric matrix

Question

I'm looking for a closed form of the following determinant $$\left|\begin {array} c 1 & \frac{1}{x_1x_2+y_1y_2} & \frac{1}{x_1x_3+y_1y_3}\\ \frac{1}{x_2x_1+y_2y_1} & 1 & \frac{1}{x_2x_3+y_2y_3} \\ \frac{1}{x_3x_1+y_3y_1} & \frac{1}{x_3x_2+y_3y_2} & 1 \end {array}\right| $$ such that $x_i^2+y_i^2=1$ for $i=1,2,3$.

I have tried several types of operations on columns and lines without success.

Answer 1

if $x_i^2 + y_i^2 = 1$ we can say that there exist a $\theta$ such that

$x_i = \cos \theta_i, y_i = \sin\theta_i$

$x_1x_2 + y_1y_2 = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 = \cos(\theta_1 - \theta_2)$

$\phi_1 = \theta_1 - \theta_2, \phi_2 = \theta_2 - \theta_3, \phi_3 = \theta_3 - \theta_1\\-\phi_3 = \phi_1 + \phi_2$

Now your matrix is

$\begin{bmatrix} 1&\sec\phi_1&\sec\phi_3\\\sec\phi_1& 1&\sec\phi_2\\\sec\phi_3&\sec\phi_2&1\end{bmatrix}$

$1 + 2\sec\phi_1\sec\phi_2\sec\phi_3 - \sec^2\phi_1 - \sec^2\phi_2-\sec^2\phi_3$