Question

The complete solution set of $\theta$ which satisfy the equation $2\tan \theta -\cot \theta =-1$ is

• A. $\{n\pi -\frac{\pi }{4}\}\cup \{m\pi +{\tan }^{-1}\left(\frac{1}{2}\right)\};m,n\in Z$
• B. $\{n\pi +\frac{\pi }{4}\}\cup \{m\pi \pm {\tan }^{-1}\left(\frac{1}{2}\right)\};m,n\in Z$
• C. $\{n\pi -\frac{\pi }{4}\},n\in Z$
• D. $\{m\pi +{\tan }^{-1}\left(\frac{1}{2}\right)\},m\in Z$

Answer 1

The correct option is A $\{n\pi -\frac{\pi }{4}\}\cup \{m\pi +{\tan }^{-1}\left(\frac{1}{2}\right)\};m,n\in Z$

$2\tan \theta -\cot \theta =-1$
For $\tan \theta ,\cot \theta$ to be defined, we get
$\theta \ne \frac{(2n+1)\pi }{2},\theta \ne n\pi$

Now,
$\Rightarrow 2\tan \theta -\frac{1}{\tan \theta }=-1\Rightarrow 2{\tan }^2\theta +\tan \theta -1=0\Rightarrow (\tan \theta +1)(2\tan \theta -1)=0\Rightarrow \tan \theta =-1,\tan \theta =\frac{1}{2}\Rightarrow \theta =n\pi -\frac{\pi }{4},\theta =m\pi +{\tan }^{-1}\frac{1}{2}\therefore \theta =\{n\pi -\frac{\pi }{4}\}\cup \{m\pi +{\tan }^{-1}\left(\frac{1}{2}\right)\};m,n\in Z$