Question

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

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

Answer 1

Answer: (B)

${\sin }^2\theta \sec \theta +\sqrt{3}\tan \theta =0$
$\Rightarrow {\sin }^2\theta \frac{1}{\cos \theta }+\sqrt{3}\frac{\sin \theta }{\cos \theta }=0$
$\Rightarrow \frac{\sin \theta }{\cos \theta }\left[\sin \theta +\sqrt{3}\right]=0$
Since , $-1\le \sin \le 1$
$\therefore \sin \theta +\sqrt{3}\ne 0\Rightarrow \frac{\sin \theta }{\cos \theta }=0$
$\Rightarrow \tan \theta =0\Rightarrow \theta =n\pi$