How to find this ?

I'm stuck at $1$ and how to $1$ to change the tangent associated with a,b or c.

I meant this ,everybody.


  • 1
    $\begingroup$ You will find this in any decent treatment of trigonometry. Did you do anything at all? $\endgroup$ – Allawonder Jun 23 '19 at 16:40
  • 5
    $\begingroup$ Welcome to stackexchange. Please edit the question to show us how you started and where you are stuck. You might begin with the identity for $\tan{A+B)$. $\endgroup$ – Ethan Bolker Jun 23 '19 at 16:40
  • $\begingroup$ Welcome to Mathematics Stack Exchange. Are you familiar with Trigonometric Addition Formulas? $\endgroup$ – J. W. Tanner Jun 23 '19 at 16:41
  • $\begingroup$ ohh thz for reply every body $\endgroup$ – Myat Linn Jun 23 '19 at 16:44
  • $\begingroup$ Do you mean this here $${\frac {\tan \left( a \right) +\tan \left( b \right) +\tan \left( y \right) -\tan \left( a \right) \tan \left( b \right) \tan \left( y \right) }{1-\tan \left( a \right) \tan \left( b \right) -\tan \left( a \right) \tan \left( y \right) -\tan \left( b \right) \tan \left( y \right) }} $$ $\endgroup$ – Dr. Sonnhard Graubner Jun 23 '19 at 17:10

Let's call $X=\tan(x)$ you have the addition formula $\tan(u+v)=\dfrac{U+V}{1-UV}$

  • call $c=a+b$ and develop $\tan(c+y)$
  • replace $C$ by $\tan(a+b)$ developpement
  • simplify $\dfrac{\frac{A+B}{1-AB}+Y}{1-\frac{A+B}{1-AB}Y}$
  • $\begingroup$ And with one more step $\frac{A+B+Y-ABY}{1-(AB+AY+BY)}.$ $\endgroup$ – Aaron Meyerowitz Jun 23 '19 at 22:03

I found the appropriate answer. I thought this is misunderstanding of english context and i don't know why? The question from GelfAndSaul Trigonometry Book is

Represent tan(a+b+c) to only representing tan(a),tan(b) and tan(c).

I have an answer but i thought that is wrong.

So this questions means only is we can neglect to 1?


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