<s>[INST] 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.
 [/INST] Just use the sum formula twice:
$$
\begin{align*}
\tan(A+B+Y)=\tan((A+B)+Y) &= \frac{\tan(A+B)+\tan Y}{1-\tan(A+B)\tan Y}\\
&=\frac{\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)+\tan Y}{1-\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)\tan Y}\\
&= \frac{(\tan A + \tan B)+ \tan Y(1-\tan A\tan B)}{(1-\tan A\tan B)-(\tan A + \tan B)\tan Y}\\
&=\frac{\tan A + \tan B + \tan Y - \tan A\tan B\tan Y}{1-\tan A\tan B-\tan A\tan Y-\tan B\tan Y}
\end{align*}
$$
</s>