Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
 A: With all these simple solutions available, how about a complex one?


*

*The argument of the product of two complex numbers is the sum of their arguments.

*The imaginary part of a complex number divided by its real part is the tangent of its argument.


Consider the product
$$
\begin{align}
&(1+i\tan(A))(1+i\tan(B))(1+i\tan(Y))\\
&=1-\tan(A)\tan(B)-\tan(B)\tan(Y)-\tan(Y)\tan(A)\\
&+i\,(\tan(A)+\tan(B)+\tan(Y)-\tan(A)\tan(B)\tan(Y))\tag{1}
\end{align}
$$
Using 1. and 2., $(1)$ says that
$$
\tan(A+B+Y)=\frac{\tan(A)+\tan(B)+\tan(Y)-\tan(A)\tan(B)\tan(Y)}{1-\tan(A)\tan(B)-\tan(B)\tan(Y)-\tan(Y)\tan(A)}\tag{2}
$$
A: We have
$$
\tan(A+B+C)=\tan(A+(B+C))=\frac{\tan A+\tan(B+C)}{1-\tan A \tan(B+C)}=
$$
$$
\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B \tan C}}{1-\tan A\frac{\tan B+\tan C}{1-\tan B\tan C}}=
\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}
$$
In the last step we multiplied the numerator and the denominator by $1-\tan B\tan C$.
A: Replacing $B$ by $B+Y$ in your expression for $\tan(A+B)$ gives
$$\dfrac{\tan A + \tan(B+Y)}{1 - \tan A \tan(B+Y)}$$
Expanding gives
$$\dfrac{\tan A + \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}{1 - \tan A \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}$$
Multiplying through by $1 - \tan B \tan Y$ gives
$$\frac{\tan A(1 - \tan B \tan Y) + \tan B + \tan Y}{1 - \tan B \tan Y - \tan A(\tan B + \tan Y)}$$
And simplifying gives you what you want.
A: Do $a=\tan A, b= \tan B$ and $y = \tan Y$. Then,
\begin{eqnarray}
\tan (A+B+Y) &=& \dfrac{\tan(A+B) + \tan(Y)}{1-\tan(A+B)y}\\
&=& \dfrac{\dfrac{a+b}{1-ab}+y}{1 - \Big(\dfrac{a+b}{1-ab}\Bigr)y}\\
&=& \dfrac{\dfrac{a+b+(1-ab)y}{1-ab}}{\dfrac{1-ab - (a+b)y}{1-ab}}\\
&=& \dfrac{a+b+y -aby}{1-ab-ay-by}.\\
\end{eqnarray}
A: Just use the sum formula twice:
$$
\begin{align*}
\tan(A+B+Y)=\tan((A+B)+Y) &= \frac{\tan(A+B)+\tan Y}{1-\tan(A+B)\tan Y}\\
&=\frac{\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)+\tan Y}{1-\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)\tan Y}\\
&= \frac{(\tan A + \tan B)+ \tan Y(1-\tan A\tan B)}{(1-\tan A\tan B)-(\tan A + \tan B)\tan Y}\\
&=\frac{\tan A + \tan B + \tan Y - \tan A\tan B\tan Y}{1-\tan A\tan B-\tan A\tan Y-\tan B\tan Y}
\end{align*}
$$
A: Just like when you use an eraser on pencil written paper stuff, erase $b$ and in its place write $B+C$ wherever it comes.
You know
$$\tan(A+b)=\frac{\tan A+\tan b}{1-\tan A\tan b}$$
It becomes:
$$\tan(A+B+C)=\frac{\tan A+\tan (B+C)}{1-\tan A\tan (B+C)}$$
Now expand and simplify... that's it!
Remember, this way you can even go to next step( for practice) to expand $\tan(A+B+C+D) !$
Suggest (for practice) use procedure in this thread outlined by robjohn 
$$ (1+i\tan(A))(1+i\tan(B))(1+i\tan(Y)) (1+i\tan(Z))$$
to appreciate how it goes in complex algebra and trig connections .
