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 \le1 $
$\therefore \, \sin \theta+ \sqrt{3} \ne0 \Rightarrow \frac{\sin \theta }{\cos \theta }=0 $
$\Rightarrow \tan\theta= 0\Rightarrow \theta =n\pi$