<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] 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 $1-\tan B\tan C$.
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