Question

The value of $\theta$ which satisfy the equation $3{\tan }^2\theta +3\tan \theta -\cot \theta =1$ (where $n\in Z$) can be

• A. $n\pi -\frac{\pi }{6}$
• B. $n\pi -\frac{\pi }{4}$
• C. $n\pi +\frac{\pi }{6}$
• D. $n\pi +\frac{\pi }{4}$

Answer 1

Answer: (A), (B), (C)

$n\pi +\frac{\pi }{6}$

$3{\tan }^2\theta +3\tan \theta -\cot \theta =1$
$\Rightarrow 3{\tan }^2\theta +3\tan \theta -\frac{1}{\tan \theta }=1$
$\Rightarrow 3{\tan }^3\theta +3{\tan }^2\theta -\tan \theta -1=0\Rightarrow (\tan \theta +1)(3{\tan }^2\theta -1)=0$
$\Rightarrow \tan \theta =-1$ or ${\tan }^2\theta =\frac{1}{3}$
$\Rightarrow \tan \theta =\tan \left(\frac{-\pi }{4}\right)\text{or}{\tan }^2\theta ={\tan }^2\left(\frac{\pi }{6}\right)$
$\Rightarrow \theta =n\pi -\frac{\pi }{4},n\pi \pm \frac{\pi }{6},n\in Z$