Finding eigenvalues of a matrix involving hyperbolic trig functions For all $t\ge 0$, we note $r=\sqrt{1+4a}$ where $a\ge 0$ is a parameter and we consider  $$
A_t=\begin{pmatrix}\left(\cosh\left(\frac{t}{2}\right)+r\;\sinh\left(\frac{t}{2}\right)\right)^2+4a\;\sinh^2\left(\frac{t}{2}\right)&4a^{1/2}r\;\sinh^2\left(\frac{t}{2}\right)\\4a^{1/2}r\;\sinh^2\left(\frac{t}{2}\right)&\left(\cosh\left(\frac{t}{2}\right)-r\;\sinh\left(\frac{t}{2}\right)\right)^2+4a\;\sinh^2\left(\frac{t}{2}\right)\end{pmatrix}
$$
I want to determine the eigenvalues of $A_t$.
$\lambda$ is an eigenvalue of $A_t$ if and only if $\det\left(A_t-\lambda\; I\right)=0$
This is what I tried to do:
I note 
$$\alpha\left(t\right)=\left(\cosh\left(\frac{t}{2}\right)+r\;\sinh\left(\frac{t}{2}\right)\right)^2+4a\;\sinh^2\left(\frac{t}{2}\right)$$ and $$\beta\left(t\right)= 4a^{1/2}r\;\sinh^2\left(\frac{t}{2}\right)$$  then
$$A_t=\begin{pmatrix}\alpha\left(t\right)&\beta\left(t\right)\\ \beta\left(t\right)&\alpha\left(t\right)-4r\;\sinh\left(\frac{t}{2}\right)\;\cosh\left(\frac{t}{2}\right)\end{pmatrix}
$$ then $$\begin{align*}
\det\left(A_t-\lambda\; I\right)&=\left(\alpha\left(t\right)-\lambda\right)\left(\alpha\left(t\right)-4r\;\sinh\left(\frac{t}{2}\right)\;\cosh\left(\frac{t}{2}\right)-\lambda\right)-\beta\left(t\right)^2\\&=\left(\alpha\left(t\right)-\lambda\right)^2-4r\;\sinh\left(\frac{t}{2}\right)\;\cosh\left(\frac{t}{2}\right)\left(\alpha\left(t\right)-\lambda\right)-\beta\left(t\right)^2
\end{align*}$$ 
and I am blocked there. Please help me. Thanks in advance.
 A: Actually you were pretty much done, you basically found the following expression
$$\begin{align*}
\det\left(A_t-\lambda\; I\right) = \left(\alpha\left(t\right)-\lambda\right)^2-\gamma(t)\left(\alpha\left(t\right)-\lambda\right)-\beta\left(t\right)^2
\end{align*}$$
with 
$$\alpha\left(t\right)=\left(\cosh\left(\frac{t}{2}\right)+r\;\sinh\left(\frac{t}{2}\right)\right)^2+4a\;\sinh^2\left(\frac{t}{2}\right),$$ 
$$\beta\left(t\right)= 4a^{1/2}r\;\sinh^2\left(\frac{t}{2}\right)$$
and
$$\gamma(t) = 4r\;\sinh\left(\frac{t}{2}\right)\;\cosh\left(\frac{t}{2}\right)$$
Which is just a quadratic polynomial of $(\alpha\left(t\right)-\lambda)$, so its solutions are
$$
\alpha\left(t\right)-\lambda = \frac{-\gamma(t) \pm \sqrt{\gamma(t)^2 + 4 \beta(t)}}{2}
$$
or equivalently
$$
\lambda  = \alpha\left(t\right) + \frac12\gamma(t) \mp \frac12\sqrt{\gamma(t)^2 + 4 \beta(t)}.
$$
A: Here is my answer please help me to correct it or to improve it if there is possibale simplifications:
We note $X(t)=\alpha(t)-\lambda$
We have 
$\begin{align*} \det(A_t-\lambda\; I)&=X(t)^2-4r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})X(t)-\beta(t)^2
 \end{align*}$
$\begin{align*}\Delta(t)&=16r^2\;sinh(\frac{t}{2})^2\;cosh(\frac{t}{2})^2+4\beta(t)^2\\&=16r^2\;sinh(\frac{t}{2})^2\;cosh(\frac{t}{2})^2+4*16ar^2sinh^4(t/2)=16r^2sinh^2(t/2)(cosh(\frac{t}{2})^2+a\;sinh^2(t/2)\ge 0\end{align*}$
$\begin{align*}X_1(t)&=\frac{4r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})-4rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sh^2(t/2)}}{2}\\&=2r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})-2rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sinh^2(t/2)}\end{align*}$
$\begin{align*}X_2(t)&=\frac{4r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})+4rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sinh^2(t/2)}}{2}\\&=2r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})+2rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sinh^2(t/2)}\end{align*}$
Then the two eigenvalues of  $A_t$ are: 
$\begin{align*}\lambda_1(t)&=\alpha(t)-X_1(t)\\&=\Big(cosh(\frac{t}{2})+r\;sinh(\frac{t}{2})\Big)^2+4a\;sinh^2(\frac{t}{2}) -2r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})+2rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sinh^2(t/2)}\end{align*}$
and
$\begin{align*}\lambda_2(t)&=\alpha(t)-X_2(t)\\&=\Big(cosh(\frac{t}{2})+r\;sinh(\frac{t}{2})\Big)^2+4a\;sinh^2(\frac{t}{2}) -2r\;sinh(\frac{t}{2})\;cosh(\frac{t}{2})-2rsinh(t/2)\sqrt{(cosh(\frac{t}{2})^2+4a\;sinh^2(t/2)}\end{align*}$
