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:

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?

  • $\begingroup$ An aside consideration: some solutions (not all, e.g. x=0 mod $\pi$) may be found using [this identity] (math.stackexchange.com/q/477364) which says that it is sufficient (but not necessary ! ) that $x,2x,3x$ are angles of a triangle ; therefore $6x=\pi (mod \pi)$ with $0<x<\pi/3$. $\endgroup$
    – Jean Marie
    Aug 1, 2016 at 16:46

4 Answers 4


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.


You concluded from

$$\dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}{1-\tan\theta\tan2\theta}=(\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$$

(by “canceling” $\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$ from both sides) that


The correct conclusion is that either $\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}=0$ or $2-\tan\theta\tan2\theta=\tan\theta\tan2\theta$. And you still have to check your solutions and keep only those for which the original equation is defined.

Canceling terms doesn’t give you an equivalent equation (one with the same solution set). If you cancel zero, you can lose solutions. If you cancel something undefined, you can introduce wrong solutions.

For example, consider the equation $$\frac{x}{1-x}=\frac{1}{1-x}.$$ If you “cancel” $1-x$ from each side, you get $x=1$, but that is not a solution, because the equation is not defined when $x=1$. And consider the equation $${x}({1-x})=2({1-x}).$$ If you “cancel” $1-x$, the only solution to what’s left is $x=2$, and you lose the other solution ($x=1$) to the equation, because for that value, you canceled zero from each side.

To summarize, if $E\cdot A=E\cdot B$, where $E$, $A$, and $B$ are expressions, solutions occur when both

  • $E\cdot A$ and $E\cdot B$ are defined
  • Either $E=0$ or $A=B$ (or both).

Set, provisionally, $x=\tan\theta$ and $y=\tan2\theta$. Then the equation becomes $$ x+y+\frac{x+y}{1-xy}=xy\frac{x+y}{1-xy} $$ Note that the expression only makes sense for $1-xy\ne0$.

After removing the denominator, we get $$ (x+y)(2-xy)=xy(x+y) $$ so this becomes $$ 2(x+y)(1-xy)=0 $$ Since $1-xy\ne0$, we remain with $x+y=0$. So $\tan\theta=-\tan2\theta$ and therefore $$ \tan\theta=\tan(\pi-2\theta) $$ Hence $$ \theta=\pi-2\theta+k\pi $$ that gives $$ \theta=\frac{\pi}{3}+k\frac{\pi}{3} $$ and this can be simplified into $\theta=k\pi/3$.


Clearly, $\tan\theta\tan2\theta\tan3\theta$ is undefined if $\theta\equiv\dfrac\pi2,\dfrac\pi4,\dfrac\pi6\pmod\pi$



$$\implies3\theta=n\pi-3\theta\iff\theta=\dfrac{n\pi}6$$ where $n$ is any integer

But $3\nmid n$ else $\tan\theta\tan2\theta\tan3\theta$ will be undefined


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