Question 634797
The key is the fact that A, B and C are angles of a triangle. So A+B+C = 180.<br>
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:
{{{tan(C) = (tan(180)-tan(A+B))/(1+tan(180)*tan(A+B))}}}
Since tan(180) = 0 this simplifies:
{{{tan(C) = (0-tan(A+B))/(1+0*tan(A+B))}}}
{{{tan(C) = (-tan(A+B))/(1+0)}}}
{{{tan(C) = (-tan(A+B))/1}}}
{{{tan(C) = -tan(A+B)}}}
Using the tan(x+y) formula on the right side we get:
{{{tan(C) = -((tan(A)+tan(B))/(1-tan(A)*tan(B)))}}}
which simplifies to
{{{tan(C) = (-tan(A)-tan(B))/(1-tan(A)*tan(B))}}}
Remember this equation. We will use it now and again later.<br>
We can substitute the above for tan(C) in the left side of the given equation (which we are trying to prove):
{{{tan(A) +tan(B) + (-tan(A)-tan(B))/(1-tan(A)*tan(B)) = tan(A)*tan(B)*tan(C)}}}<br>
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:
{{{tan(A)*((1-tan(A)*tan(B))/(1-tan(A)*tan(B))) +tan(B)((1-tan(A)*tan(B))/(1-tan(A)*tan(B))) + (-tan(A)-tan(B))/(1-tan(A)*tan(B)) = tan(A)*tan(B)*tan(C)}}}
{{{((tan(A)-tan^2(A)*tan(B))/(1-tan(A)*tan(B))) +((tan(B)-tan(A)tan^2(B))/(1-tan(A)*tan(B))) + ((-tan(A)-tan(B))/(1-tan(A)*tan(B))) = tan(A)*tan(B)*tan(C)}}}
When we add, some of the terms in the numerators cancel:
{{{((cross(tan(A))-tan^2(A)*tan(B))/(1-tan(A)*tan(B))) +((cross(tan(B))-tan(A)tan^2(B))/(1-tan(A)*tan(B))) + ((cross(-tan(A))-cross(tan(B)))/(1-tan(A)*tan(B))) = tan(A)*tan(B)*tan(C)}}}
leaving:
{{{(-tan^2(A)*tan(B)-tan(A)tan^2(B))/(1-tan(A)*tan(B))) = tan(A)*tan(B)*tan(C)}}}<br>
The right side has factors of tan(A) and tan(B). So does the left side. So lets factor them out:
{{{(tan(A)*tan(B)*(-tan(A)-tan(B))/(1-tan(A)*tan(B))) = tan(A)*tan(B)*tan(C)}}}
which can be rewritten as:
{{{tan(A)*tan(B)*((-tan(A)-tan(B))/(1-tan(A)*tan(B))) = tan(A)*tan(B)*tan(C)}}}
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):
{{{tan(A)*tan(B)*tan(C) = tan(A)*tan(B)*tan(C)}}}