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

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The discussion focuses on proving the formula tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y)tan(y-z)tan(z-x). Participants clarify that the proof relies on the assumption that A + B + C = nπ, where A, B, and C are defined as the angles x-y, y-z, and z-x, respectively. This leads to the conclusion that A + B + C equals zero, which is consistent with the formula. Confusion arises regarding the generality of the sum of angles equating to nπ, but it is emphasized that this is a specific case. The conversation highlights the importance of understanding the context of angle sums in trigonometric identities.
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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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