Because of
$\overline{\cosh(z)}=\cosh(\bar{z})$ it can be produced by
In[23]:= Simplify[TrigExpand[TrigReduce[Cosh[x + I y] Cosh[x - I y]]] /. {Sin[y]^2 -> (1 - Cos[y]^2),
Cosh[x]^2 -> (1 + Sinh[x]^2)}]
Out[23]= Cos[y]^2 + Sinh[x]^2
and then checked again by
In[37]:= FullSimplify[Cosh[x + I y] Cosh[x - I y] - (Cos[y]^2 + Sinh[x]^2)]
Out[37]= 0
B.t.w. there are equivalent answers, e.g. if one takes the other side of the basic identies in replacement
In[39]:= Simplify[TrigExpand[TrigReduce[Cosh[x + I y] Cosh[x - I y]]] /. {Cos[y]^2 -> (1 - Sin[y]^2),
Sinh[x]^2 -> (Cosh[x]^2 - 1)}]
Out[39]= Cosh[x]^2 - Sin[y]^2
In[41]:= FullSimplify[Cosh[x + I y] Cosh[x - I y] - Cosh[x]^2 + Sin[y]^2]
Out[41]= 0