SOLUTION: Hi, this question has completely stumped me. Let A,B,C be the angles of a triangle. Show that tanA+tanB+tanC=tanAtanBtanC I've tried using identities to turn it into sin/cos b

Question 634797: Hi, this question has completely stumped me.
Let A,B,C be the angles of a triangle. Show that tanA+tanB+tanC=tanAtanBtanC
I've tried using identities to turn it into sin/cos but I got stuck and couldn't go any further.
I appreciate your help,
Thankyou
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
The key is the fact that A, B and C are angles of a triangle. So A+B+C = 180.

Solving this equation for C we get:
C = 180 - (A+B)
So
tan(C) = tan(180-(A+B))
Using the tan(x-y) formula on the right wide we get:

Since tan(180) = 0 this simplifies:

Using the tan(x+y) formula on the right side we get:

which simplifies to

Remember this equation. We will use it now and again later.

We can substitute the above for tan(C) in the left side of the given equation (which we are trying to prove):

The right side is one term (of three factors). So it will help if we make the left side one term, too. So we'll get common denominators and add:

When we add, some of the terms in the numerators cancel:

leaving:

The right side has factors of tan(A) and tan(B). So does the left side. So lets factor them out:

which can be rewritten as:

And if you look at the third factor and at the equaiton I have asked you to remember, you will see that they are the same. So the third factor is tan(C):

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