# 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
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!

##### Well-known member
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
"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
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
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 $$\pi$$.