Question

If $\tan \theta =-\frac{1}{\sqrt{3}}$ then the general solution of the equation

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

Answer 1

Answer: (D)

$\tan \theta =-\frac{1}{\sqrt{3}}=\tan \left(-\frac{\pi }{6}\right)$
$\therefore \theta =n\pi -\frac{\pi }{6}$