Question

General value of $\theta$ satisfying the equation ${\tan }^2\theta +\sec 2\theta =1$ is

• A. $m\pi ,n\pi +\frac{\pi }{3}$
• B. $m\pi ,n\pi \pm \frac{\pi }{3}$
• C. $m\pi ,n\pi \pm \frac{\pi }{6}$
• D. None of these

Answer 1

Answer: (B)

$m\pi ,n\pi \pm \frac{\pi }{3}$

By using, $\sec 2\theta =\frac{1}{\cos 2\theta }=\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }$
We can write the given equation as
${\tan }^2\theta +\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }$
$\Rightarrow {\tan }^2\theta +\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=1$
$\Rightarrow {\tan }^2\theta (1-{\tan }^2\theta )+1+{\tan }^2\theta =1-{\tan }^2\theta$
$\Rightarrow 3{\tan }^2\theta -{\tan }^4\theta =0$
$\Rightarrow {\tan }^2\theta (3-{\tan }^2\theta )=0$
$\Rightarrow \tan \theta =0or\tan \theta =\pm \sqrt{3}$
Now, $\tan \theta =0\Rightarrow \theta =m\pi$, where $m$ is an integer and $\tan \theta =\pm \sqrt{3}\tan [\pm \frac{\pi }{3}]$
$\Rightarrow \theta =n\pi \pm \frac{\pi }{3}$ where $n$ is an integer
Thus, $\theta =m\pi ,n\pi \pm \frac{\pi }{3}$