Proof of $\mathrm{tr}(\gamma^{5}\gamma^\mu\gamma^\nu)=0$ Using $\gamma^{5}\gamma^\mu=-\gamma^\mu\gamma^{5}$ and $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ I obtain
\begin{equation}\tag{1}
T_{\mu\nu}:=\mathrm{tr}(\gamma^{5}\gamma^\mu\gamma^\nu)=-\mathrm{tr}(\gamma^\mu\gamma^{5}\gamma^\nu)=-\mathrm{tr}(\gamma^{5}\gamma^\nu\gamma^\mu)=-T_{\nu\mu}.
\end{equation}
However, I need to prove
\begin{equation}
T_{\mu\nu}=-T_{\mu\nu}.
\end{equation}
Wikipedia gives a hint:

Simply add two factors of $\gamma^{\alpha}$, with $\alpha$ different from $\mu$ and $\nu$. Anticommute three times instead of once, picking up three minus signs, and cycle using the cyclic property of the trace.

Maybe I don't understand the hint, in any case, by inserting $\gamma^{\alpha}\gamma^{\alpha}=-1$ I simply obtain equation $(1)$.
 A: Following the given hint, we note that $\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha} = \mathbb{I}$, for $\alpha=0,1,2,3$.
Thus, we can write
$$\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5) =\mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5\,\gamma^{\alpha}\,\gamma^{\alpha}) \quad. $$
Further note that for $\alpha\neq \mu$ and $\alpha\neq \nu$ it holds that $$\{\gamma^{\mu},\gamma^{\alpha}\} = \{\gamma^{\nu},\gamma^{\alpha}\} = \{\gamma^{5},\gamma^{\alpha}\} = 0 \quad .$$
By making use of these anti-commutation relations, we can rearrange the first equation to
\begin{align} \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5\,\gamma^{\alpha}\,\gamma^{\alpha}) &= - \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^{\alpha}\,\gamma^{5}\,\gamma^{\alpha}) \\
&= \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\alpha}\,\gamma^{\nu}\,\gamma^{5}\,\gamma^{\alpha}) \\
&= -\mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^{5}\,\gamma^{\alpha}) \quad .
\end{align}
The cyclic properties of the trace then eventually yield
$$\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5)  = -\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5) \quad . $$
