Question

If ${\sin }^2\theta \sec \theta +\sqrt{3}\tan \theta =0$, then $\theta$ is

Answer 1

${\sin }^2\theta \sec \theta +\sqrt{3}\tan \theta =0$

${\sin }^2\theta /\cos \theta +\sqrt{3}\tan \theta =0$

$\sin \theta \tan \theta +\sqrt{3}\tan \theta =0$

$\tan \theta (\sin \theta +\sqrt{3})=0$

So, putting

$\tan \theta =0$ and $\sin \theta +\sqrt{3}=0$

as we can see that $\sin \theta =-\sqrt{3}$ is not possible.

so, by $\tan \theta =0$, we get $\theta =n\pi$