So the ideas is writing $t$ in terms of $\eta$ and put it into the $sinh(Ht)$.
By using $dta=d\eta$ we can write
$d\eta = \frac{dt}{da}$ hence $$\eta = \int \frac{dt}{sinh(Ht)} = \frac{1}{H}ln(tanh(\frac{Ht}{2}))$$
Thus
$$\eta = \frac{ln(tanh(Ht/2)}{H}$$
if you take the the inverse of it you will see that it's
$$t = \frac{2arctanh(e^{H\eta})}{H}$$
Then I inserted this into the $sinh(Ht)$ to get a function in terms of $\eta$.
So $$a(\eta) = sinh(2arctanh(e^{H\eta}))$$
Now at this point I kinda got stuck so I ask for a help at math stack exchange and here is the link
https://math.stackexchange.com/questions/3607046/simplifying-sinh-left2-operatornamearctanh-leftehx-right-right
And they proved that
$$a(\eta) = sinh(2arctanh(e^{H\eta})) = -csch(H\eta)$$