MCQ
The general solution of the trigonometric equation $\tan ^2 \theta=1$ is ....
- A$\theta=n \pi \pm \frac{\pi}{3}, n \in Z$
- B$\theta=n \pi \pm \frac{\pi}{6}, n \in Z$
- C$\theta=n \pi \pm \frac{\pi}{4}, n \in Z$
- D$\theta=n \pi, n \in Z$
$\tan ^2 \theta=1=\tan ^2\left(\frac{\pi}{4}\right)$
$\tan ^2 \theta=\tan ^2 \alpha \Rightarrow \theta=n \pi \pm \alpha$
$\therefore \theta=n \pi \pm \frac{\pi}{4}$
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