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?
Best Answer
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.