Question

The general solutions of the equation (\tan^2 \theta + \sec 2\theta = 1) are

• A. (n\pi, n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z})
• B. (n\pi, n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z})
• C. (\frac{n\pi}{4}, \frac{n\pi}{4} \pm \frac{\pi}{3}, n \in \mathbb{Z})
• D. (n\pi, n\pi \pm \frac{\pi}{6}, n \in \mathbb{Z})

Hint:

General solution for $\tan^2 \theta = \tan^2 \alpha$ is $\theta = n\pi \pm \alpha$.

Answer 1

The Correct Option is A