Short answer: it depends on the impedance of the load attached to the secondary coil.
A perfect transformer can be modeled as a pair of inductors with mutual inductance $M$ and self-inductances $L_1$ and $L_2$, with $M^2 = L_1 L_2$. Denote the voltage across coil 1 and the current through it as $V_1(t) = \tilde{V}_1 e^{i \omega t}$ and $I_1(t) = \tilde{I}_1 e^{i \omega t}$, and similarly for coil 2. Assume that we hook up a load of impedance $Z$ across coil 2, so that $\tilde{V}_2 = Z \tilde{I}_2$. Then from Kirchoff's laws we end up with the equations
\begin{align*}
-i \omega L_1 \tilde{I}_1 - i \omega M \tilde{I}_2 &= \tilde{V}_1 && \text{(coil 1)} \\
- i \omega M \tilde{I}_1 - i \omega L_2 \tilde{I}_2 &= \tilde{V}_2 = Z \tilde{I}_2 && \text{(coil 2)}
\end{align*}
This can be solved for $\tilde{I}_1$ and $\tilde{I}_2$, to yield
\begin{align*}
\tilde{I}_1 &= -\frac{i Z - L_2 \omega}{L_1 Z \omega} \tilde{V}_1 & \tilde{I}_2 &= - \frac{M}{L_1 Z} \tilde{V}_1
\end{align*}
In the limit that the load impedance is large compared to the inductance of the secondary coil, we recover the result that $\tilde{I}_1$ is 90° out of phase with $\tilde{V}_2$, and so very little power is dissipated in this case. (It's not hard to see that the power dissipated in coil 2, which is something like $\frac{1}{2} \Re ( Z \tilde{I}_2^* \tilde{I}_2)$, is also very small in this limit, so that's a good sanity check.) However, if the load impedance is comparable to the impedance of the secondary coil, then $\tilde{I}_1$ and $\tilde{V}_1$ will have a more complicated phase relationship.
