# Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$

Fernando begs for help on this most difficult trig 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.

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(-1) I understand, it is very comfortable when other people put in order your post. But now its time to learn some $\TeX$. – Norbert Aug 1 '12 at 16:43
I just dont understand latex its not my fault. – Fernando Martinez Aug 1 '12 at 16:44
Take a look at this giudeline – Norbert Aug 1 '12 at 16:45
@FernandoMartinez Let this be an opportunity to learn it. It is not hard. – Tibor Aug 1 '12 at 16:45
Can you remove the thumbs down at least it breaking my....heart – Fernando Martinez Aug 1 '12 at 16:55

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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 It makes sense thanks. – Fernando Martinez Aug 1 '12 at 16:45

With all these simple solutions available, how about a complex one?

1. The argument of the product of two complex numbers is the sum of their arguments.
2. 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 $$\tan(A+B+Y)=\frac{\tan(A)+\tan(B)+\tan(Y)-\tan(A)\tan(B)\tan(Y)}{1-\tan(A)\tan(B)-\tan(B)\tan(Y)-\tan(Y)\tan(A)}\tag{2}$$

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 +1) that's the solution, if you ask me :) – Valentin Aug 3 '12 at 20:39

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.

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 Very help answer thanks. – Fernando Martinez Aug 1 '12 at 16:45

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}

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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*}

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 Although all responses are good Fernando likes this one especially. – Fernando Martinez Aug 1 '12 at 17:15

I want to give simplest one. as we know $$A+B+C=\pi$$ $$A+B=\pi-c$$ $$tan(A+B)=tan(\pi-c)$$ $$\frac {tan(A)+tan(B)}{1-tan(A)tan(B)}=-tan(C)$$ $$tan(A)+tan(B)=-tan(C)+tan(A)tan(B)tan(C)$$ $$tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C)$$ proved

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 can somebody tell me why this solution has downvote. – iostream007 May 10 at 4:02