<s>[INST] Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$ 
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?
 [/INST] When you cancel out the terms from LHS and RHS, you drop the solutions when these terms are 0 or do not exist (because the denominator is 0). A trivial example would the $\theta=0$, which certainly is a solution, but you did not find it because of the canceled terms.
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