Question

If equation in variable $\theta$, $3\tan (\theta -\alpha )=\tan (\theta +\alpha )$, (where $\alpha$ is constant), has no real solution, then $\alpha$ can be (wherever $\tan (\theta -\alpha )$ and $\tan (\theta +\alpha )$ both are defined)

• A. $\frac{\pi }{15}$
• B. $\frac{5\pi }{18}$
• C. $\frac{5\pi }{12}$
• D. $\frac{17\pi }{18}$

Answer 1

The correct option is B $\frac{5\pi }{18}$

$\frac{\tan (\theta +\alpha )}{\tan (\theta -\alpha )}=3$
Using componendo and dividendo, we get
$\sin 2\theta =2\sin 2\alpha$
This equation has no solution if $|\sin 2\alpha |>\frac{1}{2}$
$\Rightarrow 2\alpha \in (\frac{\pi }{6},\frac{5\pi }{6})\cup (\frac{7\pi }{6},\frac{11\pi }{6})$
$\text{i.e.},\alpha \in (\frac{\pi }{12},\frac{5\pi }{12})\cup (\frac{7\pi }{12},\frac{11\pi }{12})$
$\therefore$ The possible value of $\alpha$ is $\frac{5\pi }{18}$ .