A problem is NP-hard iff its complement is coNP-complete
This argument is given for any complexity class.

If $C$ is a complexity class (for your question NP) and $L$ is $C-Hard$ (an 'identical' argument can be for complete languages):

If $A \in coC$ so $\bar A \in C$  so $ \bar A \le L$: ($\phi$ is the Karp-reduction for that matter)

$ \exists \phi.x\in \bar A \iff \phi(x)\in L $

Since it's iff the negation is: 

$ \exists \phi.x\notin \bar A \iff \phi(x) \notin L$ 

Hence, (by definition of complementary)

$ \exists \phi.x\in A \iff \phi(x) \in \bar L$ 

$A \le \bar L \implies \bar L \in coC-Hard$

Negating the answer of an NP Turing machine won't help you to decide a language in coNP (except for trivial cases like language in P), since by definition an NP Turing machine accepts iff there exist an accepting branch in its' computation. 

For simplicity, assume your NP Turing machine always has a rejecting branch in every computation (we can assume it for every NP Turing machine, why?), negating the answer will result in the trivial language $\Sigma $* every-time.