Question

If $\tan \theta =-\frac{1}{\sqrt{3}},$ then the value of $\theta \in [0,2\pi ]$ for which $\frac{\cos \theta -{\cos }^3\theta }{\tan \theta +2}$ is always positive is

• A. $\frac{4\pi }{3}$
• B. $-\frac{\pi }{6}$
• C. $\frac{\pi }{6}$
• D. $\frac{11\pi }{6}$

Answer 1

The correct option is D $\frac{11\pi }{6}$

$\frac{\cos \theta -{\cos }^3\theta }{\tan \theta +2}=\frac{\cos \theta (1-{\cos }^2\theta )}{\tan \theta +2}=\frac{\cos \theta {\sin }^2\theta }{\tan \theta +2}$
$\because \tan \theta +2$ is positive, $\cos \theta {\sin }^2\theta$ will be positive iff $\cos \theta$ is positive.
So, $\cos \theta >0$ and $\tan \theta <0$
$\Rightarrow \theta \in 4^{th}$ quadrant.
$\therefore \tan \theta =-\frac{1}{\sqrt{3}}$ $\Rightarrow \theta =2\pi -\pi /6=\frac{11\pi }{6}$