Question

If $2ta{n}^2\theta =se{c}^2\theta$, then the general solution of $\theta$

• A. $n\pi +\frac{\pi }{4},n\in I$
• B. $n\pi -\frac{\pi }{4},n\in I$
• C. $n\pi \pm \frac{\pi }{4},n\in I$
• D. $2n\pi \pm \frac{\pi }{4},n\in I$

Answer 1

The correct option is C $n\pi \pm \frac{\pi }{4},n\in I$

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

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

$\Rightarrow {\tan }^2\theta =1$

$\Rightarrow {\tan }^2\theta ={\tan }^2\frac{\pi }{4}$

$\Rightarrow \tan \theta =\pm \tan \frac{\pi }{4}$

So,the general solution is

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