Ans: \((A, C)\)
\(z=\sin \theta-i \cos \theta\)
$$
\begin{aligned}
& \cos \left(\theta-\frac{\pi}{2}\right)+i \sin \left(\theta-\frac{\pi}{2}\right) \\
=& e^{i\left(\theta-\frac{\pi}{2}\right)}
\end{aligned}
$$
$$
\begin{aligned}
&\text { so, } z^{n}=e^{\left(n \theta-\frac{n \pi}{2}\right)}=\cos \left(n \theta-\frac{n \pi}{2}\right)-i \sin \left(n \theta-\frac{n \pi}{2}\right) \\
&\frac{1}{z^{n}}=e^{i\left(\frac{n \pi}{2}-n \theta\right)}=\cos \left(n \theta-\frac{n \pi}{2}\right)-i \sin \left(n \theta-\frac{n \pi}{2}\right), \text { so } z^{n}+\frac{1}{z^{n}}=2 \cos \left(n \theta-\frac{n \pi}{2}\right)=2 \cos \left(\frac{n \pi}{2}-n \theta\right) \\
&z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)(C)
\end{aligned}
$$
(A)
