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If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
Wiki User
∙ 11y ago
tan (A-B) + tan (B-C) + tan (C-A)=0
tan (A-B) + tan (B-C) - tan (A-C)=0
tan (A-B) + tan (B-C) = tan (A-C)
(A-B) + (B-C) = A-C
So we can solve
tan (A-B) + tan (B-C) = tan (A-C)
by first solving
tan x + tan y = tan (x+y)
and then substituting x = A-B and y = B-C.
tan (x+y) = (tan x + tan y)/(1 - tan x tan y)
So tan x + tan y = (tan x + tan y)/(1 - tan x tan y)
(tan x + tan y)tan x tan y = 0
So, tan x = 0 or tan y = 0 or tan x = - tan y
tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C)
tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B)
A, B and C are all angles of a triangle, so are all in the range (0, pi).
So A-B and B-C are in the range (- pi, pi).
At this point I sketched a graph of y = tan x (- pi < x < pi)
By inspection I can see that:
A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi
A = B or B = C or A = C or A = C +/- pi
But A and C are both in the range (0, pi) so A = C +/- pi has no solution
So A = B or B = C or A = C
A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
Wiki User
∙ 11y ago
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