Trying to solve the energy levels of a spin 1/2 particle in a one-dimensional box using Dirac Equation

Question

I was studying the problem I asked above in the title and found the article P Alberto et al 1996 Eur. J. Phys. 17 19. The wave function inside the walls is: $$ \psi(z)=B\ exp(ikz) \left[\begin{array}{c} \chi\\ \frac{\hbar k c}{mc^2+E}\sigma_z\chi \end{array} \right] + C\ exp(-ikz) \left[\begin{array}{c} \chi\\ \frac{-\hbar k c}{mc^2+E}\sigma_z\chi \end{array} \right] $$ The boundary condition is that the flux of probability is zero at the walls: $$ \pm (i) \gamma^{z} \psi=\psi $$ Where the $\pm$ sign corresponds to $z=0$ and $z=L$, respectively and $\gamma^z$ is the Dirac gamma matrix. In $z=0$ we found: $$ C=B\frac{iP-1}{iP+1} $$ Where P is: $$ P=\frac{k\hbar c}{E+mc^2} $$ If we apply the same boundary condition in $z=L$, replacing the relation between B and C previously found, we reach to: $$ \tan(kL)=\frac{2P}{P^2 -1} $$ Then, the article said that, using the expression for P in the equation above, we reach to: $$ \tan(kL)=\frac{2P}{P^2 -1}=-\frac{\hbar k}{mc} $$ I didn't understand. Following the steps I was able to reach the first equal sign from the equation above, but the second sign is the problem. As you can see, the RHS of the equation I showed doesn't have a dependence on the eigenvalue $E$. If some of you could help me I'll thank you a lot!