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

[Elissa89]

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]

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]

"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 = $n\pi$

 

[Evgeny.Makarov]

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 $\tan A+ \tan B + \tan C = \tan A \tan B \tan C$.

 

[HOI]

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.