If $\theta$ and $\alpha$ are not odd multiples of $\frac{\pi}{2}$,then $\tan \theta = \tan \alpha$ implies the general solution is
- A$\theta = \alpha + \frac{n \pi}{2}, n \in Z$
- B$\theta = \alpha + \frac{3 n \pi}{2}, n \in Z$
- C$\theta = n \pi + \alpha, n \in Z$
- D$\theta = \frac{n \pi}{4} + \alpha, n \in Z$
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