Solving for transmission coefficient in the finite square well

Consider a finite square well of depth $V_0$ and which extends from $-a$ to $a$. For $|x|>a$, $V=0$. The wavefunction ansatz one can propose for an incoming wave from the left $Ae^{ikx}$ is:
$$ psi = Ae^{ikx} + Be^{-ikx}, x<a $$

$$ psi = Ce^{ik_2 x} + Be^{-ik_2 x}, |x|<a $$

$$ psi = Fe^{ikx}, x>a $$
Where $k=frac{2mE}{hbar^2}$ and $k_2=frac{2mE+V_0}{hbar^2}$ can be obtained from the Schrodinger Equation. Then, if we define the transmission coefficient to be:
$$ T = frac{F^2}{A^2}$$
One should be able to find the value of this coefficient if $F$ is written in terms of $A$. We can apply boundary conditions to do so: $psi$ and $psi’$ must be continuous at $-a$ and at $a$, so:
$$ Ae^{-ika}+Be^{ika} = Ce^{ik_2 a}+De^{ik_2}a $$
$$ -ik(Ae^{-ika}-Be^{ika}) = -ik_2(Ce^{-ik_2 a}-De^{ik_2a}) $$
$$ Ce^{ik_2a}+De^{-ik_2a} = Fe^{ika} $$
$$ ik_2(Ce^{ik_2a}-De^{-ik_2a}) = ikFe^{ika} $$
Now, I have seen derivations of $T$ where $psi$ is taken to be $Ccos(k_2x)+Dsin(k_2x)$, but it should be the same here and I am not being able to eliminate the $C$ and $D$ to get equations only in $B$,$F$ and $A$. Am I not seeing a way to do this, are the boundary conditions wrong or is the ansatz wrong (I have been told that either way the value of $T$ should be the same). Thanks!