$$sech^{-1}(z)=\log(\frac{1}{z}+(\frac{1}{z^2}-1)^\frac{1}{2})$$
my attempt at the moment is the following
First recall from definition that $sech(z)=\frac{2}{e^z+e^-z}$, now let $x=\frac{2}{e^z+e^-z}$ $=\frac{2}{e^z+e^-z}(\frac{e^z}{e^z})$ $=\frac{2e^z}{e^{2z} +1}$ $\implies x(e^{2z})+x-2e^z=0$, then by the quadratic formula and some rearranging i arrive at $z=\log(\frac{1}{x}+(\frac{1}{x^2}-1)^{\frac{1}{2}})$ which is clearly incorrect, i was wondering if anyone could supply me with a better solution as i'm stuck, Thanks.
we have $$y=\frac{2}{e^z+e^{-z}}$$ multiplying numerator and denominator by $e^z$ we have $$e^{2z}-\frac{2}{y}e^z+1=0$$ solving this we get $$e^z_{1,2}=\frac{1}{y}\pm\sqrt{\frac{1}{y^2}-1}$$ Can you go on?