Question

Solve for the general solution for the given equation: $2\tan \theta -\cot \theta =-1$

• A. $\theta \in \{n\pi +\alpha \}\bigcup \{n\pi -\frac{3\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$
• B. $\theta \in \{n\pi +\alpha \}\bigcup \{n\pi +\frac{\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$
• C. $\theta \in \{n\pi +\alpha \}\bigcup \{n\pi +\frac{3\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$
• D. $\theta \in \{n\pi -\alpha \}\bigcup \{n\pi +\frac{3\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$

Answer 1

The correct option is C $\theta \in \{n\pi +\alpha \}\bigcup \{n\pi +\frac{3\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$

Given: $2\tan \theta -\cot \theta =-1$
Let's convert the equation in terms of $\tan x$ only.

$\Rightarrow 2\tan \theta -\frac{1}{\tan \theta }=-1$ for $\tan \theta \ne 0$
$\Rightarrow 2{\tan }^2\theta -1=-\tan \theta$
$\Rightarrow 2{\tan }^2\theta +\tan \theta -1=0$

This is a quadratic equation in $\tan x.$
$\Rightarrow 2{\tan }^2\theta +(2-1)\tan \theta -1=0$
$\Rightarrow 2{\tan }^2\theta +2\tan \theta -\tan \theta -1=0$
$\Rightarrow 2\tan \theta (\tan \theta +1)-(\tan \theta +1)=0$
$\Rightarrow (2\tan \theta -1)(\tan \theta +1)=0$

Here, either $2\tan \theta -1=0$ or $\tan \theta +1=0$

For the first case, $\tan \theta =\frac{1}{2}$
We know, $\tan \theta \in R.$
$\therefore$ There exist an angle $\alpha$ such that $\tan \alpha =\frac{1}{2}.$
The general solution, say A, for $\tan \theta =\tan \alpha$ is $\theta =n\pi +\alpha ,n\in Z.$

For the second case, $\tan \theta =-1$
We know, $\tan \frac{\pi }{4}=|-1|.$
Also, $\tan \theta$ is negative in the second and fourth quadrant.

$\therefore$ The possible angles are $\tan (-\frac{\pi }{4})=-1;$
$\tan (\pi -\frac{\pi }{4})=-1\Rightarrow \tan \frac{3\pi }{4}=-1$
$\tan (2\pi -\frac{\pi }{4})=-1\Rightarrow \tan \frac{7\pi }{4}=-1$

Considering $\tan \theta =\tan \frac{3\pi }{4},$ the general solution, say B, is $\theta =n\pi +\frac{3\pi }{4},n\in Z.$

The complete general solution is $A\cup B.$
$\theta \in \{n\pi +\alpha \}\bigcup \{n\pi +\frac{3\pi }{4}\},n\in Z,$ where $\tan \alpha =\frac{1}{2}.$