im$12%. By considering the forms of the effective couplings of the Higgses to the bottom quark, we can determine if this enhancement translates to a nonnegligible correction. The effective couplings are given by [@Carena:1999py] \_b\^h &=& (1 - )\
\_b\^H &=& (1 + )\
\_b\^A &=& (1 - ) , where $g^{h,H,A}_b$ are the tree level couplings. In the decoupling limit, $\tana {\ensuremath{\rightarrow}}-\cotb$ and we obtain \_b\^h &=& g\^h\_b\
\_b\^H &=& (1 - )\
\_b\^A &=& (1 - ) . We therefore only need to determine the extent to which $\Delta_1$ affects the size of the factor $(1+\Delta_b)^{-1}$. From \[Eq:deltas\], the factor may be written as = \_[12]{} . Let us define $\delta_2\equiv (1+\Delta_2\tanb)^{-1}$ and $\delta_\Phi$ to be the relative change between ignoring $\Delta_1$ and including it, \_ . By setting $\Delta_1=0.12$, $\delta_\Phi$ can be plotted as a function of $\Delta_2\tanb$ as shown in \[fig:deltas\]. For positive values of $\Delta_2\tanb$, the relative change is never more than 6%. Unless $\Delta_2\tanb$ is $\mathcal{O}(1)$, the relative correction to the heavy Higgs couplings is only a few percent. The effect of including $\Delta_1$ can be more drastic if $\Delta_2\tanb$ is negative. As $\Delta_2\tanb$ approaches $-\mathcal{O}(1)$, the relative change increases quickly to the nearly the same magnitude. Such large, negative values of $\Delta_2\tanb$ may be a more extreme case however. For most values of $\Delta_2\tanb$ obtained in the parameter scan $(<40\%)$, the relative change is again only a few percent. Thus unless the magnitude of the $\tanb$-enhanced corrections to the bottom quark mass are $\mathcal{O}(1)$ it is safe to neglect the $\Delta_1$ correction to the couplings of the bottom quark with the heavy Higgses.[^5] Note that in calculating the $B_1^{\widetilde{g}}$ contribution to $\Delta_1$ we take $Q=M_Z$. If the scale is chosen to be higher, then $B_1^{\widetilde{g}}$ would be smaller and the relative change, $\delta_\Phi$, would be more suppressed.

![The plot shows the relative size of