Question

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

Answer 1

${\tan }^2\theta +\sec 2\theta =1\Rightarrow {\tan }^2\theta +\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=1\Rightarrow {\tan }^2\theta -{\tan }^4\theta +1+{\tan }^2\theta =1-{\tan }^2\theta$
$\Rightarrow {\tan }^4\theta -3{\tan }^2\theta =0\Rightarrow {\tan }^2\theta ({\tan }^2\theta -3)=0$

$\text{If}{\tan }^2\theta =0\Rightarrow {\tan }^2\theta ={\tan }^{2}0\Rightarrow \theta =m\pi ,m\in Z....\left(i\right)$
and

${\tan }^2\theta =3\Rightarrow {\tan }^2\theta ={\tan }^2\frac{\pi }{3}\Rightarrow \theta =n\pi \pm \frac{\pi }{3},n\in Z....\left(ii\right)$