|And the answer is ...|
Message #15 Posted by Dave Shaffer on 28 May 2002, 9:45 p.m.,
in response to message #14 by Steve (Australia)
With the a+b+c = 180 degrees condition, things are trivial.
Recall Steve's initial fact:
tan(a+b) = (tan a + tan b)/(1 - (tan a)*(tan b))
which can be looked up in your favorite book of trig identities. Note, also, that tan(-a) = -tan a .
Then, with the 180 degree condition, we need to determine whether
tan a + tan b + tan(180-a-b) = (tan a)*(tan b)*tan(180-a-b)
Note, also, that tan(180-x) = -tan x (which can be derived from the sum formula above, realizing that tan 180 = 0).
Then we get tan(180-a-b) = -tan(a+b)
This relation greatly simplifies our task!
To keep from typing "tan" all over the place, I let
tan a = a, tan b = b, and tan c = c everywhere below. Since only tangents appear, there is no loss of meaning.
Then, our expression to check becomes (with the 180 degree
identity included, which puts in some minus signs):
a + b - tan(a+b) =? -a * b * tan(a+b)
where tan(a+b) has its normal meaning
(i.e. NOT tan(tan a + tan b)) !!
Now, expand tan(a+b) to get
a + b - [(a + b)/(1-ab)] =? -a * b * [(a + b)/(1-ab)]
where a, b, c now all mean their respective tangents.
Multiply this expression through by the (1-ab) denominator to get
(1-ab)(a+b) - a - b =? -a * b * (a+b)
a + b - aab - abb -a - b =? -aab - abb
where I have elected to write aa (and bb) instead of a^2
(a-squared) for a times a.
The a and b terms cancel on the left side, leaving us with an identity: -aab -abb = -aab -abb
So, indeed, if the sum of the angles equals 180 degrees, the sum and product of the three tangent terms ARE EQUAL.