Proving Addition/Subtraction Formulas - Tan(x-y)+Tan(y-z)+Tan(z-x)

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Discussion Overview

The discussion centers on the mathematical identity involving the tangent function: tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y)tan(y-z)tan(z-x). Participants are exploring the proof of this identity, which involves the properties of angles and the tangent function, while expressing confusion over the steps and assumptions involved.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants express confusion over the proof and the steps involved, indicating that previous attempts have been unclear or overly complicated.
  • One participant suggests that the identity follows from the condition A + B + C = nπ, leading to the relationship tan A + tan B + tan C = tan A tan B tan C.
  • Another participant questions the assumption that A + B + C = nπ for arbitrary angles, seeking clarification on this point.
  • A later reply clarifies that the assumption is made for the sake of the proof, not as a general statement about arbitrary angles.
  • It is noted that if A = x - y, B = y - z, and C = z - x, then A + B + C = 0, which corresponds to 0 times π, thus supporting the initial claim.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the assumptions regarding the angles A, B, and C. There is ongoing debate about the validity of the proof and the conditions under which the identity holds.

Contextual Notes

The discussion highlights the dependence on the assumption that A + B + C = nπ, which is not universally applicable without specific conditions. The steps in the proof are also noted to be complex and potentially confusing for those not familiar with the underlying concepts.

Elissa89
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tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

I have redone this problem two or three times and all the steps just make my head spin. I've tried looking up tutorials online but they introduce things into the problem that we haven't been taught yet and that just confuses me more. Help!
 
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Elissa89 said:
tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

I have redone this problem two or three times and all the steps just make my head spin. I've tried looking up tutorials online but they introduce things into the problem that we haven't been taught yet and that just confuses me more. Help!

this follows from $A+B+C=n\pi=>\tan\, A+ \tan \, B + \tan\, C = \tan\, A \tan \, B \tan\, C$

if you want a proof of the above

$A+B+C=n\pi => A+B=n\pi-C$
take tan on both sides getting

$\frac{\tan\, A + \tan\, B}{1-\tan\, A \tan \, B} = -\tan\, C$
Or
$\tan\, A + \tan\, B = - \tan\, C + \tan\, A \tan \, B\tan\, C$
or
$\tan\, A + \tan\, B + \tan\, C = \tan\, A \tan \, B\tan\, C$

as (x-y) + (y-z) + (z-x) = 0 we get the result
 
"this follows from
[FONT=MathJax_Math-italic]A[FONT=MathJax_Main]+[FONT=MathJax_Math-italic]B[FONT=MathJax_Main]+[FONT=MathJax_Math-italic]C[FONT=MathJax_Main]=[FONT=MathJax_Math-italic]n[FONT=MathJax_Math-italic]π"

I don't understand. If A, B and C are arbitrary angles how can we know that their sum = $n\pi$
 
DavidCampen said:
I don't understand. If A, B and C are arbitrary angles how can we know that their sum = $n\pi$

Nobody claims that $A+B+C=n\pi$ in general. On the contrary, we assume it. Kaliprasad meant that your equation follows from the fact that if $A+B+C=n\pi$, then $\tan A+ \tan B + \tan C = \tan A \tan B \tan C$.
 
In your original post you were asking about tan(x-y)+tan(y-z)+tan(z-x). If A= x- y, B= y- z, and C= z- x, then A+ B+ C= x- y+ y- z+ z- x= 0 which is 0 times \pi.
 

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