Welcome to our community

Be a part of something great, join today!

proving with addition and subtraction formulas: tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

Elissa89

Member
Oct 19, 2017
52
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!
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
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
 

DavidCampen

Member
Apr 4, 2014
64
"this follows from
[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]C[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]π[/FONT]"

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

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
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 $\tan A+ \tan B + \tan C = \tan A \tan B \tan C$.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
371
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 [tex]\pi[/tex].