Question

If $2{\tan }^2\theta ={\sec }^2\theta$, then the general solution of $\theta$ is

• A. $n\pi +\frac{\pi }{4}\left(nϵI\right)$
• B. $n\pi -\frac{\pi }{4}\left(nϵI\right)$
• C. $n\pi \pm \frac{\pi }{4}\left(nϵI\right)$
• D. $2n\pi \pm \frac{\pi }{4}\left(nϵI\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $n\pi +\frac{\pi }{4}\left(nϵI\right)$
B $n\pi -\frac{\pi }{4}\left(nϵI\right)$

Given $2{\tan }^2\theta ={\text{sec}}^2\theta$

$⟹2{\tan }^2\theta =1+{\tan }^2\theta$

$⟹{\tan }^2\theta =1$

So $\theta =n\pi \pm \frac{\pi }{4},n\in I$