Question

General value of $\theta$ satisfying the equation ${\tan }^2\theta +sec2\theta =1$ is ________?

Answer 1

Ans. $\theta =n\pi ,n\pi \pm \pi /3$
$t^2+\frac{1+t^2}{1-t^2}=1$ where $t=\tan \theta$
$\therefore t^2(t^2-3)=0$ $\therefore \tan \theta =0,\pm \sqrt{3}$
$\therefore \theta =n\pi ,n\pi \pm \pi /3$.