I needed to solve the following equation: $$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first: $$\begin{split} \tan\theta + \tan 2\theta+\tan 3\theta &= (\tan\theta + \tan 2\theta) + \dfrac{\tan\theta + \tan 2\theta} {1-\tan\theta\tan2\theta} \\ &= \dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)} {1-\tan\theta\tan2\theta} \end{split}$$
And, RHS yields $$\begin{split} \tan\theta\tan2\theta\tan3\theta &= (\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta} {1-\tan\theta\tan2\theta} \end{split}$$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$$
\begin{split}
2-\tan\theta\tan2\theta &= \tan\theta\tan2\theta\\
\tan\theta\tan2\theta &= 1,
\end{split}$$
which can be further reduced as:
$$\tan^2\theta=\frac{1}{3}\implies\tan\theta=\pm\frac{1}{\sqrt3}$$
Now, we can yield the general solution of this equation:
$\theta=n\pi\pm\dfrac{\pi}{6},n\in Z$. But, setting $\theta=\dfrac{\pi}{6}$ in the original equation is giving one term $\tan\dfrac{\pi}{2}$, which is not defined.
What is the problem in this computation?