Conserving algebraic quantities I have the relations 
$$r_0^{\pm} = \frac{1}{2}\left (\frac{\hat \rho_0\zeta_1^{1/2}}{\hat \rho_1\zeta_0^{1/2}} \pm \frac{\zeta_0^{1/2}}{\zeta_1^{1/2}} \right )$$
$$r_1^{\pm} = \frac{1}{2}\left (\frac{\hat \rho_1\zeta_2^{1/2}}{\hat \rho_2\zeta_1^{1/2}} \pm \frac{\zeta_1^{1/2}}{\zeta_2^{1/2}}\right )$$
$$T_{0,1} = \frac{1}{r_{0,1}^{+}}$$
$$R_{0,1}= -\frac{r_{0,1}^{-}}{r_{0,1}^{+}}$$
and I'm trying to show that 
$$R_{0,1}^2+T_{0,1}^2 = 1.$$
if $\hat \rho_0 = \hat \rho_2$ and $\zeta_0=\zeta_2.$ 
So far I've failed to show this. When I compute the above, I get 
$$R_0^2+T_0^2 = \frac{\hat \rho_1^2 \zeta_0^2 + 4\hat \rho_1^2 \zeta_0 \zeta_1 - 2 \hat \rho_1 \hat \rho_0 \zeta_0 \zeta_1 + \hat \rho_0^2 \zeta_1^2}{(\hat \rho_1 \zeta_0 + \hat \rho_0 \zeta_1)^2}$$
If I could somehow find a way to factor this to achieve $1$, it would be great. Could anyone cast an eye over this to see if I'm not spotting the obvious? 
Thanks in advance.

For $i\in\{0,1\}$ you want to show that $R_i^2+T_i^2=1$, or equivalently
$$(r_i^+)^2-(r_i^-)^2=1.$$
For the sake of legibility let $a_k:=\hat{\rho}_k$ and $b_k:=\zeta_k^{1/2}$. Then for $i=0$ we have
$$r_0^{\pm}=\frac12\left(\frac{a_0b_1}{a_1b_0}\pm \frac{b_0}{b_1}\right)
=\frac{a_0b_1^2\pm a_1b_0^2}{2a_1b_0b_1},$$
and so the standard identity $(x+y)^2-(x-y)^2=4xy$ shows that
\begin{eqnarray*}
(r_0^+)^2-(r_0^-)^2
&=&\frac{(a_0b_1^2+a_1b_0^2)^2}{4a_1^2b_0^2b_1^2}
-\frac{(a_0b_1^2-a_1b_0^2)^2}{4a_1^2b_0^2b_1^2}\\
&=&4\frac{a_0a_1b_0^2b_1^2}{4a_1^2b_0^2b_1^2}=\frac{a_0}{a_1},
\end{eqnarray*}
and so the identity holds if and only if $a_0=a_1$. Because $a_2=a_0$ and $b_2=b_0$, the exact same proof works for $i=1$ by changing the indices; change all the $0$'s to $1$'s and all the $1$'s to $0$'s.
