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I found this question in a maths worksheet of trigonometry (kinda odd, right?), but I dont know how to figure it out.

If $\displaystyle x+y+z=xyz$, find $\displaystyle\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$

First I thought of taking x,y and z as $\displaystyle \tan A, \tan B,$ and $\tan C,$ making $A+B+C=\pi$, but couldnt solve ahead. Any solution not involving trigonometry would do as well. I also think that this question does not even relate to trigo.....or does it?

lab bhattacharjee
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Rohinb97
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3 Answers3

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We can not make $A+B+C=\pi$

as $\displaystyle \tan(A+B+C)=\frac{\sum\tan A-\tan A\tan B\tan C}{1-\sum \tan A\tan B}$

$\displaystyle\sum\tan A=\tan A\tan B\tan C\implies \tan(A+B+C)=0\implies A+B+C=n\pi$ where $n$ is any integer [Clearly, you have taken a special value$(1)$ of $n$]

Now for any integer $m,$ $\displaystyle m(A+B+C)=mn\pi\implies \tan(mA+mB+mC)=\tan mn\pi=0$

$\displaystyle\implies\sum\tan mA=\tan mA\tan mB\tan mC$

Now set $m=3$ and apply $\tan3x$ formula

lab bhattacharjee
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  • Since tanA +tanB+tanC=tanAtanBtanC, then A+B+C=$\pi$, that's all I said. And I think in the 4th line you meant $n=1$. But it's more generalised and I see your point. Nice! – Rohinb97 Mar 03 '14 at 13:39
  • @Rohinb97, thanks for your observation. I was thinking to rectify when I'd just shut down and then forgot forever – lab bhattacharjee Mar 03 '14 at 14:55
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Hint: You are in the right track. Now observe that $\tan (3A+3B+3C)=0$ and hence $\tan 3A+\tan 3B+\tan 3C=\tan 3A\tan 3B \tan 3C$

Samrat Mukhopadhyay
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Case I: At least one of $x^2$, $y^2$ and $z^2$ is equal to $1/3$. This case is possible, e.g., $x=\frac{\sqrt{3}}{3}=-y$ and $z=0$. In this case, the formula $\displaystyle \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ does not make any sense.

Case II: None of $x^2$, $y^2$ and $z^2$ is equal to $1/3$ and $xyz=0$. For example, we say $x=-y$ and $z=0$, which implies that \begin{eqnarray*} \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}+\frac{-3x+x^3}{1-3x^2}+0=0. \end{eqnarray*}

Case III: None of $x^2$, $y^2$ and $z^2$ is equal to $1/3$ and $xyz\neq0$. Take $x=\tan A$, $y=\tan B$ and $z=\tan C$ for some $A,B,C\in(-\pi/2,\pi/2)$, since $x+y+z=xyz$, then $xy,xz,yz\neq 1$, which implies that $A+B$, $A+C$ and $B+C$ can not be $\pm\pi/2$.

Claim I: $A+B+C\neq\pm\pi/2$.

If not, then $\tan(A+B)=\cot C$, that is, $\displaystyle\frac{x+y}{1-xy}=\frac{1}{z}$, that is, $(x+y)z=1-xy$. Hence $(x+y)z^2=z-xyz=z-(x+y+z)=-(x+y)$, that is, $x+y=0$. Since $x+y+z=xyz$ and $z\neq0$, we get $xy=1$, contradiction.

In summary, we can define $\tan(3A)$, $\tan(3B)$, $\tan(3C)$ and $\tan(A+B+C)$. Since $\displaystyle \tan(3\theta)=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$, then we have \begin{eqnarray*} \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\tan(3A)+\tan(3B)+\tan(3C). \end{eqnarray*}

Notice that \begin{eqnarray*} \tan(A+B+C)&=&\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C}\\ &=&\frac{x+y+z-xyz}{1-xy-xz-yz}=0. \end{eqnarray*}

Then there exists some $m\in\{-1,0,1\}$ such that $A+B+C=m\pi$. Hence $\tan(A+B)+\tan C=0$, which implies that $\tan^2(A+B)\neq1/3$. Hence we can define $\tan(3A+3B)$. So we have \begin{eqnarray*} 0&=&\frac{\tan(3A)+\tan(3B)+\tan(3C)-\tan(3A)\tan(3B)\tan(3C)}{1-\tan(3A)\tan(3B)-\tan(3A)\tan(3C)-\tan(3B)\tan(3C)}. \end{eqnarray*}

Then we get \begin{eqnarray*} \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\tan(3A)+\tan(3B)+\tan(3C)=\tan(3A)\tan(3B)\tan(3C). \end{eqnarray*}

m-agag2016
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  • Taking this many cases is not required, according to me, but still, I got what you said (wrote?). Thanks! – Rohinb97 Mar 04 '14 at 07:53