Question

The general value of $\theta$ satisfying the equation $ta{n}^2\theta +sec2\theta =1$ is___

Answer 1

$ta{n}^2\theta +sec2\theta =1$
$t^2+\frac{1+t^2}{1-t^2}=1,wheret=tan\theta$
or $t^2(t^2-3)=0$
or $tan\theta =0,\pm \sqrt{3}$
$\Rightarrow \theta =n\pi$ and $\theta =n\pi \pm \frac{\pi }{3},n\in Z$