# Finding an expression for $\tan (a+ b + y)$ which involves only $\tan a, \tan b,$ and $\tan y$.

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.

$$\frac{tan(a)+tan(b)+tan(y)−tan(a)tan(b)tan(y)}{1−tan(a)tan(b)−tan(a)tan(y)−tan(b)tan(y)}$$

• You will find this in any decent treatment of trigonometry. Did you do anything at all? – Allawonder Jun 23 '19 at 16:40
• 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 $\tan{A+B)$. – Ethan Bolker Jun 23 '19 at 16:40
• Welcome to Mathematics Stack Exchange. Are you familiar with Trigonometric Addition Formulas? – J. W. Tanner Jun 23 '19 at 16:41
• ohh thz for reply every body – Myat Linn Jun 23 '19 at 16:44
• Do you mean this here $${\frac {\tan \left( a \right) +\tan \left( b \right) +\tan \left( y \right) -\tan \left( a \right) \tan \left( b \right) \tan \left( y \right) }{1-\tan \left( a \right) \tan \left( b \right) -\tan \left( a \right) \tan \left( y \right) -\tan \left( b \right) \tan \left( y \right) }}$$ – 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 $$\dfrac{\frac{A+B}{1-AB}+Y}{1-\frac{A+B}{1-AB}Y}$$
• And with one more step $\frac{A+B+Y-ABY}{1-(AB+AY+BY)}.$ – 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?