Arctanh to exp: Prove two equations are equivalent For some peace of mind in a project, I am trying to prove two equations are somewhat equivalent. I have these two equations.
$$
i_{1} = \frac{-I_M}{2}\frac{2i_2\left(1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)+i_e\left(-1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)}{2i_2\left(-1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)+i_e\left(1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)}
$$
$$
i_1 = \frac{I_M}{2}\frac{\left(-1\pm e^{\left(\mp2\frac{i_2R_E + V_T\operatorname{Arctanh}\left[\frac{i_2}{i_e}\right]}{V_T}\right)}\right)}{\left(1\pm e^{\left(\mp 2\frac{i_2R_E + V_T\operatorname{Arctanh}\left[\frac{i_2}{i_e}\right]}{V_T}\right)}\right)}
$$
I know these equations are equivalent because their Taylor series coefficients are pretty much the same except for one factor of 2 on the denominator. I have probably made a mistake with the formation of one of the two equations above, but I can't find it, and I think if I convert the top one to a form similar to the bottom one I can 'debug' the error.
I also know the Arctanh to exponential identity gives a very similar form to the top one, (seen here 5th one down) but I don't know how to convert between the two with the extra variables in there.
I have tried many times but this is abit beyond me. Does someone want to have a go a converting Eq. 1 to Eq. 2 or vice versa? Even if it's not close, it might give me some insight.
Thanks a lot.
Edit: added missing $V_T$ and minus sign into eq. 2.
 A: There are way too many distracting symbols floating around. To simplify, I ignore the "$i_1$"s and "$I_M/2$"s. Also, I define
$$a := i_2 \qquad b := i_e \qquad x := \frac{i_2 R_E}{V_T} \qquad y := \operatorname{atanh}\frac{i_2}{i_e} = \operatorname{atanh}\frac{a}{b}$$
Moreover, I assert that, in the second equation, "$\pm$" is "$+$", and "$\mp$" is "$-$"; and, in the first equation, I suppress the factors of "$2$", which I believe are in error.
Then the question becomes
$$- \frac{a \left(e^{2x} + 1 \right) + b \left( e^{2x}-1 \right)}{a \left( e^{2x}-1\right) + b \left( e^{2x}+1 \right) } \quad \stackrel{?}{=} \quad 
-\frac{1-e^{-2(x+y)}}{1+e^{-2(x+y)}}$$
or, more simply,
$$\frac{a \cosh x + b \sinh x}{a \sinh x + b \cosh x} \quad \stackrel{?}{=} \quad
\frac{\sinh(x+y)}{\cosh(x+y)} = \tanh(x+y)$$
(Here, I multiplied-through the numerator and denominator on the left by $\frac{1}{2}e^{-x}$, and on the right by $\frac{1}{2}e^{(x+y)}$, and then invoked the definitions of $\sinh$ and $\cosh$.)
Now, dividing-through the right by $b\cosh x$, and recalling the definition of $y$, gives
$$\frac{\frac{a}{b} + \tanh x}{\frac{a}{b}\tanh x+1} = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y} \quad \stackrel{\ddot{\smile}}{=} \quad \tanh(x+y)$$ 
