Eigenvalues of a particular tridiagonal matrix with a variation on one of its diagonal entries

Question

I am trying to obtain the eigenvalues of this particular $n\times n$ tridiagonal matrix \begin{eqnarray} A & = & \begin{bmatrix} 1+a^{2} & -a & 0 & \ldots & 0 & 0 \\ -a & 1+a^{2} & -a & \ldots & 0 & 0 \\ 0 & -a & 1+a^{2} & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \vdots & 1+a^{2} & -a \\ 0 & 0 & 0 & \vdots & -a & 1 \end{bmatrix} \end{eqnarray} and my concern is the bottom right entry with a 1 instead of $1+a^{2}$. Is it possible to work out these eigenvalues? When the bottom right entry is $1+a^{2}$ it is known that the eigenvalues are given by $\lambda_{s}=1+a^{2}-2 \ a \ Cos(s\pi/(n+1))$ but I have been unable to confirm a similar outcome for this variation.