Question

The general solution of the trigonometric equation ${\tan }^2\theta =1$ is ______

• A. $\theta =n\pi \pm \frac{\pi }{4},n\in z$
• B. $\theta =n\pi ,n\in z$
• C. $\theta =n\pi \pm \frac{\pi }{3},n\in z$
• D. $\theta =n\pi \pm \frac{\pi }{6},n$

Answer 1

The correct option is A $\theta =n\pi \pm \frac{\pi }{4},n\in z$

${\tan }^2\theta =1={\tan }^2\left(\frac{\pi }{4}\right)$

${\tan }^2\theta ={\tan }^2\alpha \Rightarrow \theta =n\pi \pm \alpha$

$\theta =n\pi \pm \frac{\pi }{4}$