Q) If $\theta_{1},\theta_{2},\theta_{3}$ are three distinct values lying in $[0,3\pi)$ for which $\tan \theta =\frac{7}{10}$ then $$\left[\tan\frac{\theta_{1}}{3}\tan \frac{\theta_{2}}{3}+\tan\frac{\theta_{2}}{3}\tan \frac{\theta_{3}}{3}+\tan\frac{\theta_{3}}{3}\tan \frac{\theta_{1}}{3}\right]$$ is =___ (where $[.]$ denotes G.I.F) ?
My Attempts
My basic analysis was that $$0<\theta_{1}<\frac{\pi}{2}$$ $$\pi<\theta_{2}<\frac{3\pi}{2}$$ $$2\pi<\theta_{1}<\frac{5\pi}{2}$$ (I might be wrong here)
Multiplied by $\frac{1}{3}$ and applied $\tan$ on LHS and RHS .
$$0<\tan\frac{\theta_{1}}{3}<\frac{1}{\sqrt{3}}$$ $$\sqrt{3}<\tan\frac{\theta_{2}}{3}<\infty$$ $$-\sqrt{3}<\tan\frac{\theta_{1}}{3}<-\frac{1}{\sqrt{3}}$$
Now we know that $\tan$ has fundamental period $\pi$ so we have
$$\theta_{1}$$ $$\theta_{2}=\pi+\theta_{1}$$ $$\theta_{3}=2\pi+\theta_{1}$$
Now if my procedure so far is accurate how do i use this to effectively solve the question ?
HINT:
Use $\tan3A=\dfrac{3\tan A-\tan^3A}{1-3\tan^2A}$ in $\dfrac7{10}=\tan\theta$ to form a Cubic equation in $\tan\dfrac\theta3$
Now use Vieta's formula