Question

If $\tan \theta +\tan 2\theta +\tan \theta \tan 2\theta =1$ then general value of $\theta$ is-

• A. $n\pi ;n\in I$
• B. $n\pi \pm \frac{\pi }{3};n\in I$
• C. $\frac{n\pi }{3}+\frac{\pi }{12},n\in I$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{n\pi }{3}+\frac{\pi }{12},n\in I$

Given $\tan \theta +\tan 2\theta +\tan \theta \tan 2\theta =1$

We know

$\tan (\theta +2)=\frac{\tan \theta \to \tan 2}{1-\tan \theta \tan \alpha }$

then of $\tan (\theta +\alpha )=1$

then we will have $\tan \theta +\tan \alpha +\tan \theta \tan \alpha =1$

for $\theta =\theta$ and $\alpha =2\theta$ and $\tan (\theta +\alpha )=\tan (\theta +2\theta )=1$

we have

$\tan \theta =1=\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }$

let

$\Rightarrow$ Which given the required expression

$\therefore \tan 3\theta =1$ we know $\tan A=1$

when $A=\frac{\pi }{4}n\in I$

or $n\pi +\frac{\pi }{4}$

$\therefore 3\theta =n\pi +\frac{\pi }{4}$

$\Rightarrow \theta =n\frac{\pi }{3}+\frac{\pi }{12}$

$n\in I$