Question

The general solution of equation ${\sin }^2\theta \sec \theta +\sqrt{3}\tan \theta =0$ is

• A. $\theta =n\pi +{(-1)}^{n+1}\frac{\pi }{3}$
• B. $\theta =n\pi$
• C. $\theta =n\pi +{(-1)}^{n+1}\frac{\pi }{6}$
• D. $\theta =\frac{n\pi }{2}$

Answer 1

Answer: (B)

$\theta =n\pi$

Given

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

${\sin }^2\theta \frac{1}{\cos \theta }+\sqrt{3}\frac{\sin \theta }{\cos \theta }=0$

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

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

$\sin \theta =0$

$(\sin \theta \ne -\sqrt{3})$

$\theta =n\pi$