The correct answer is $\boxed{\text{C) }a_n=0\text{ for }n\text{ even and }b_n=0\text{ for }n\text{ odd}}$.
A periodic signal $x(t)$ has a trigonometric Fourier series expansion
$$x(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t) \right)$$
where $\omega_0 = 2\pi/T$ is the fundamental frequency of the signal and $T$ is the period of the signal.
If $x(t) = -x(-t)$, then $x(t)$ is an odd function. If $x(t) = -x\left(\frac{t-\pi}{\omega_0}\right)$, then $x(t)$ is a periodic function with period $\frac{2\pi}{\omega_0}$.
The Fourier coefficients of an odd function are all zero for even values of $n$, and the Fourier coefficients of a periodic function with period $\frac{2\pi}{\omega_0}$ are all zero for odd values of $n$.
Therefore, if $x(t) = -x(-t) = -x\left(\frac{t-\pi}{\omega_0}\right)$, then $a_n=0$ for $n$ even and $b_n=0$ for $n$ odd.