Assume $G$ be the fundamental group of a closed orientable hyperbolic 3-manifold. Let 
$G\_{1} = \langle a\_{1},...,a\_{k} \rangle$ be a free subgroup of $G$, and let $G\_{2}=\langle a\_{k+1} \rangle$ be a free cyclic subgroup of $G$. 


My question is:
If the set $\{a\_{1},...,a\_{k}, a\_{k+1}\}$ generates $G$, 
does the rank of $G$ then satisfy $r(G)\geq k$?



a question on rank of fundamental group
Suppose that $M$ fibers over circle; $G=\pi\_1(M)\cong P\_g \rtimes <a>$, where $P\_g$ is the geneus $g$ surface group. Then for each sufficiently large $k$, you can find a free rank $k$ subgroup $F\_k<P\_g<G$ such that $F\_k$, and $a$ generate $G$. 


Edit: Let $D$ be a generating set of $P\_g$. Partition $D$ in two disjoint nonempty subsets $D=B\sqcup C$. Then for all sufficiently large $n$, the group generated by
$$
a^nBa^{-n}, a^{-n}Ca^n
$$
is free (this follows from Klein combination and the fact that conjugation by $a$ corresponds to a pseudo-Anosov homeomorphism of the surface). This is your subgroup $F\_k$. 


Thus, the answer to your question is negative. In view of the virtual fibration conjecture, the same will be probably true for all compact hyperbolic 3-manifolds. 


