Question

General value of $\theta$ satisfying the eqation $ta{n}^2\theta +sec2\theta =1$ is

• A. $m\pi ,n\pi \pm \frac{\pi }{3},m,n\in I$
• B. $m\pi ,n\pi \pm \frac{\pi }{4},m,n\in I$
• C. $m\pi ,n\pi \pm \frac{\pi }{6},m,n\in I$
• D. $m\pi ,n\pi \pm \frac{\pi }{8},m,n\in I$

Answer 1

The correct option is A $m\pi ,n\pi \pm \frac{\pi }{3},m,n\in I$

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

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

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

$\Rightarrow 3{\tan }^2\theta -{\tan }^4\theta =0\Rightarrow {\tan }^2\theta (3-{\tan }^2\theta )=0$

$\Rightarrow \tan \theta =0$ or $\tan \theta =\pm \sqrt{3}=\pm \tan \frac{\pi }{3}$

$\Rightarrow \theta =m\pi$ or $\theta =n\pi \pm \frac{\pi }{3}$