Question

If $\theta_{1}, \theta_{2}, \theta_{3}$ are three values lying in $[0,3 \pi)$ for which $\tan \theta=\lambda$, then the value of $\left|\tan \left(\frac{\theta_{1}}{3}\right) \tan \left(\frac{\theta_{2}}{3}\right)+\tan \left(\frac{\theta_{2}}{3}\right) \tan \left(\frac{\theta_{3}}{3}\right)+\tan \left(\frac{\theta_{3}}{3}\right) \tan \left(\frac{\theta_{1}}{3}\right)\right|$ is

• A. 1 /3
• B. 1
• C. 3
• D. none of these

Answer 1

Answer: (C)

$\tan \theta=\frac{3 \tan \frac{\theta}{3}-\tan ^{3} \frac{\theta}{3}}{1-3 \tan ^{2} \frac{\theta}{3}}=\lambda$
$\Rightarrow \tan ^{3}\left(\frac{\theta}{3}\right)-3 \lambda \tan ^{2}\left(\frac{\theta}{3}\right)-3 \tan \frac{\theta}{2}+\lambda=0$
Above equation has roots $\tan \frac{\theta_{1}}{3}, \tan \frac{\theta_{2}}{3}, \tan \frac{\theta_{3}}{3}$
$\therefore \left|\sum \tan \left(\frac{\theta_{1}}{3}\right) \tan \left(\frac{\theta_{2}}{3}\right)\right|=|-3|=3$