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 $\tan \frac{{\theta }_1}{3},\tan \frac{{\theta }_2}{3},\tan \frac{{\theta }_3}{3}$
$\therefore \left|\sum \tan \left(\frac{{\theta }_1}{3}\right)\tan \left(\frac{{\theta }_2}{3}\right)\right|=|-3|=3$