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

If $\theta$ and $\alpha$ are not odd multiples of $\frac{\pi}{2}$ then $\tan \theta=\tan \alpha$ implies principal solution is
  1. $\quad \theta=\alpha+\frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathbb{Z}$
  2. $\quad \theta=\alpha+\frac{3 \mathrm{n} \pi}{2}, \mathrm{n} \in \mathbb{Z}$
  3. $\quad \theta=\mathrm{n} \pi+\alpha, \mathrm{n} \in \mathbb{Z}$
  4. $\theta=\frac{n \pi}{4}+\alpha, n \in \mathbb{Z}$

Solution

$\begin{aligned} & \tan \theta=\tan \alpha \\ \therefore \quad & \theta=n \pi+\alpha, n \in Z\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

Practice more Trigonometric Functions questions on Aicharya