If $\theta$ and $\alpha$ are not odd multiples of $\frac{\pi}{2}$ then $\tan \theta=\tan \alpha$ implies principal solution is

• A

  $\quad \theta=\alpha+\frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathbb{Z}$
• B

  $\quad \theta=\alpha+\frac{3 \mathrm{n} \pi}{2}, \mathrm{n} \in \mathbb{Z}$
• Correct answer
  C

  $\quad \theta=\mathrm{n} \pi+\alpha, \mathrm{n} \in \mathbb{Z}$
• D

  $\theta=\frac{n \pi}{4}+\alpha, n \in \mathbb{Z}$

Hence, the correct answer is option: 'C'.