In fact, I came across the following statement: $$ \frac{\sinh ix}{\sinh ix+i\cosh ix}\simeq\sinh x\simeq x. $$ Unfortunately, the claim by itself is incomplete, as it is not clear what $x$ is. (Is it real/complex? Is it large/small?)
A likely argumentation is the following: Since $$ \frac{\sin x}{\sin x+\cos x}\simeq\sin x\simeq x\quad\text{for $x$ sufficiently small} $$ and $$ \frac{\sinh ix}{\sinh ix+i\cosh ix}=\frac{\sin x}{\sin x+\cos x}, $$ we have $$ \frac{\sinh ix}{\sinh ix+i\cosh ix}\simeq\sin x\simeq x. $$ But $\sinh x\simeq\sin x$ for $x$ sufficiently small, which yields the statement.
I would like to provide a more detailed verification (e.g. using series expansion), especially to see if under the condition that $x$ be sufficiently small the approximation will indeed be valid. Any help is welcome, thanks.
$\sinh x=\dfrac {x^1}{1!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots$,
so asymptotically $\sinh x\sim x$ as $x\to0$ in $\mathbb R$ or $\mathbb C$.