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.

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.
