Why less gauge fixing conditions in Faddeev Popov method? According to Dirac theory of constraints systems, to study the dynamics of gauge invariant observables, we can fix the gauge freedom by fixing it using gauge fixing conditions which are equal to number of first class constraints (even though independent parameters of gauge transformation is equal to number of primary first class constraints). So why is the case that in Yang-Mills theory we have two first class constraints but we use only one gauge fixing condition like axial gauge or Lorenz gauge in Faddeev-Popov method? 
According to Rothe and Rothe book, phase space path integral for constraint systems is
$$Z = \int dqdp \Pi_{t, r}\delta(\phi^{1}_{r})\Pi_{t, r}\delta(\xi_{r})\det\{\xi_{\alpha}, \phi^{1}_{\beta}\}\exp\left\{\int dt(qp - H_{0})\right\}$$
(we can also extend this to second-class constraints using Senjanovic path integrals)
Should we not need two gauge fixing conditions with property that Poisson bracket with gauge fixing conditions and first class constraints matrix is non-singular i.e. $\det\{\xi_{\alpha}, \phi^{1}_{\beta}\}=0$ so that we can use above path integral.
I am unable to find where my reasoning is going wrong with this.
 A: First, some clarifications.

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*It does not make sense to talk about "primary first class constraints." You refer to "primary" and "secondary" constraints while going through the Dirac procedure. After the procedure terminates, there is no distinction between primary and secondary constraints; indeed the difference depends on how you do the calculation (for example, all constraints are "primary" if you start with the full Hamiltonian with all necessary constraints included). It is only at this point, after having derived the full set of constraints, that one can distinguish between first class and second class constraints. In other words: before you know all the constraints, you don't know whether a given Poisson bracket vanishes on the constraint surface, so you can't say whether a constraint is first or second class.


*$SU(N)$ Yang-Mills theory has $N^2-1$ first class constraints (ie, the number of generators of the $SU(N)$ algebra), not 2.
With that in mind... fixing Lorentz gauge for Yang-Mills $\partial_\mu A^\mu_a=0$ does in fact impose a set of $N^2-1$ constraints on the $N^2-1$ gauge fields $A_\mu^a$. This makes sense, since there is one gauge field per generator, and one first class constraint per generator.
The counting isn't quite enough though as there is still a residual gauge symmetry even after fixing Lorentz gauge. One way to phrase the problem is that there is a pure-gauge longitudinal mode, which does not decouple from the other degrees of freedom even after fixing Lorentz gauge. Keeping track of how this pure gauge degree of freedom drops out of physical observables in a systematic way requires heavy machinery, such as BRST symmetry.
