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)

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