prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$. We have
$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$
$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$
$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$
and we have to prove that
$$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$
squaring (1) & (2), and adding them we have
$$x^2 + y^2 - z^2 = -2xy \cos(\theta - \phi)$$
multiplying (1) & (3), we have
$$x^2 + y^2 - z^2 = -xy (\cos\theta \sec\phi + \cos\phi \sec\theta)$$
from here I can't move forward
help me
 A: multiplying (2) & (3), we have
$$x^2\tan\theta+y^2\tan\phi+xy\left(\sin\theta \sec\phi+\sin\phi\sec\theta \right)=z^2\tan\psi$$
Therefore in simplest terms $\tan\theta=1$, $\tan\phi=1$ and $\tan\psi=1$ whilst at the same time 
$\cos(\theta - \phi) =-1$ from squaring (1) and (2).
One solution that meets these constraints is $\theta=\frac{\pi}{4}$, $\phi=-\frac{3\pi}{4}$ and $\psi=-\frac{3\pi}{4}$ which just indicates a fixed phase relationship between $\theta$, $\phi$ and $\psi$ I think. 
If there is a shared angle of rotation, $\alpha$, the actual solution will be in the form $\theta=\frac{\pi}{4}+\alpha$, $\phi=-\frac{3\pi}{4}+\alpha$ and $\psi=-\frac{3\pi}{4}+\alpha$.
A: We denote $a,b,c$ the angles $\theta, \phi, \psi$. The condition is
$$\begin{bmatrix} \cos a &\cos b &\cos c\\\sin a &\sin b &\sin c\\\sec a &\sec b &\sec c\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ The determinant of the matrix is equal to
$$\frac{\sin(b-a)}{\cos c}+\frac{\sin(a-c)}{\cos b}+\frac{\sin(c-b)}{
\cos a}$$ 
This determinant is in general non-zero as it is easy to verify.
Therefore the solution of the linear system above has a unique solution which is the obvious one $x=y=z=0$ in which case the equality to prove is trivially true.
The conclusion is that it remains to add some condition on the angles.
