0

I need help with the following problem - a hint of how to approach it would be fine:

Given $a$, $b$, and $c$ are real numbers and $a+b+c=abc$, find $$\tan^{-1}(a)+\tan^{-1}(b)+\tan^{-1}(c).$$

guest
  • 114
  • $$\small\tan^{-1}a + \tan^{-1}b + \tan^{-1}c = \begin{cases} \pi, & 1 < ab+bc+ca \land a + b + c > 0\ 0, & 1 > ab+bc+ca\ -\pi, & 1 < ab+bc+ca \land a + b + c < 0\end{cases}$$ See this for argument. Please note that for real $a,b,c$ which satisfies $a+b+c = abc$, it is impossible for $ab+bc+ca = 1$. – achille hui Sep 21 '19 at 04:34

4 Answers4

1

By the trigonometric identity $$ \tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}. $$ let $\alpha$, $\beta$, $\gamma$ be respectively the arctangents of $a$, $b$, and $c$ so that $$ \frac{a+b+c-abc}{1-ab-ac-bc} = \tan\Big(\arctan(a)+\arctan(b)+\arctan(c)\Big) $$ then since $a+b+c=abc$ we have that the numerator of the above equation is zero $$\tan\Big(\arctan(a)+\arctan(b)+\arctan(c)\Big)=0$$ from which we know the tangent function is zero at integer multiples of $\pi$. However, we shouldn't accept all integer values of $\pi$ since $\arctan(\theta):\mathbb R \to (-\frac{\pi}{2},\frac{\pi}{2})$ and therefore the range of $\arctan(\theta)$ is restricted to $(-\frac{\pi}{2},\frac{\pi}{2})$. Following the technique shown inside this question, we may conclude

$$\arctan(a)+\arctan(b)+\arctan(c) = \begin{cases} \pi, & 1 < ab+bc+ca ~~\text{and}~~ a+b+c > 0\\ 0, & 1 > ab+bc+ca\\ -\pi & 1 < ab+bc+ca ~~\text{and}~~ a+b+c < 0 \end{cases} $$

Axion004
  • 10,056
  • It's not $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (Instead of $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$)? – guest Sep 21 '19 at 22:42
  • It is not $[ -\frac{\pi}{2} , \frac{\pi}{2}]$. See here and remember that the tangent function has vertical asymptotes at integer multiples of $\frac{\pi}{2}$. – Axion004 Sep 21 '19 at 22:53
0

Let $\tan^{-1}a=A,a=\tan A$ etc.

For nonzero finite $a,b,c$

Observe that $-\pi/2<A,B,C<\pi/2\implies?<A+B+C<?$

$-c=(a+b)(1-ab)$

$\tan(A+B)=-\tan C=\tan(-C)$

$A+B=n\pi+(-C)$ where $n$ is any integer

lab bhattacharjee
  • 274,582
0

Let a=tanA , b= tanB and c= tanC $\\$ Since tanA + tan B + tan C = tanA.tanB.tanC $\\$ Therefore. A+B+C= nπ , for n to be any integrr. And for inverse to define $tan^{-1}x$ lie in interval (-π/2,π/2) $\\$ Hence A+B+C = -π or π

mathsdiscussion.com
  • 1,235
0

If $A+B+C=n\pi, n \in I ,$ then $$\tan A+ \tan B + \tan C= \tan A \tan B \tan C$$ So if $\tan A=a, \tan B=b, \tan C=c$, then $$\tan^{-1} a+\tan^{-1}b.+\tan^{-1} c= n \pi, n\in I.$$

Proof: Let $$A+B+C=n \pi \Rightarrow A+B =n \pi -C \Rightarrow \tan(A+B)=\tan (n \pi-C) \Rightarrow \frac{\tan A + \tan B}{1-\tan A \tan B}=-\tan C \Rightarrow \tan A + \tan B+ \tan c= \tan A \tan B \tan C.$$

Interestingly, when a=1, b=2, c=3, then $\tan^{-1} 1+ \tan^{-1} 2 + \tan^{-1} 3= \pi$ is well known.

Z Ahmed
  • 43,235
  • Can you elaborate a bit on how you got $A+B+C=n\pi?$ – guest Sep 21 '19 at 04:36
  • I understand from the previous posts how we have the substitution $a=\tan A, b=\tan B, c=\tan C,$ and that it follows that $$\tan^{-1}(a)+\tan^{-1}(b)+\tan^{-1}(c)=A+B+C,$$ but I need help on the $n\pi, n\in \mathbb{Z}^+$ part. – guest Sep 21 '19 at 04:41
  • It isn't true for all $n\pi, n\in\mathbb Z^+$. I have adjusted my answer. – Axion004 Sep 21 '19 at 05:01
  • @guest please see my edit, $n$ ia any integer it need not be positive. – Z Ahmed Sep 21 '19 at 05:31