Question

prove that

tan (A-B) +tan(B-A)+ tan(C-A) = tan(A-B) tan(B-C)t an(C-A)

Answer 1

We have to prove : tan(A - B) + tan(B - C) + tan(C - A) = tan(A - B).tan(B - C).tan(C - A)

we know the formula ,
Tan( x + y + z) = {tanx + tany + tanz - tanx.tany.tanz}/{1 - tanx.tany - tany.tanz - tanz.tanx } use this here,

(A - B) + (B - C) + (C - A) = 0
taking tan both sides,
Tan{(A - B) + (B - C) + (C - A)} = tan0 = 0
⇒ {tan(A - B) + tan(B - C) + tan(C - A) - tan(A - B).tan(B - C).tan(C - A)}/{1 - tan(A - B).tan(B - C) - tan(B - C).tan(C - A) - tan (C - A).tan(A - B)} = 0
⇒tan(A - B) + tan(B - C) + tan(C - A) - tan(A - B).tan(B -C).tan(C - A) = 0
⇒tan(A - B) + tan(B - C) + tan(C - A) = tan(A - B).tan(B - C).tan(C - A)
Hence, proved//

Answer 2

Seems like the question is wrong. The question should be tan(A - B)+tan(B-C)+tan(C-A) = tan(A-B)tan(B-C)tan(C-A). To solve this question need a formula, tan (A + B + C) = tan A + tan B + tan C - tan A tan B tan C/ 1 - tan A tan B- tan C tan A - tan B tan C. ---------(1)

(A-B)+(B-C)+(C-A) = 0
or, tan{(A-B)+(B-C)+(C-A)} = 0 [ tan0 = sin0/cos0 = 0/1 = 0 ]
now, compare (A-B) = a, (B-C) = b, (C-A) = c then you can find it more easier to solve.
or, tan(A-B)+tan(B-C)+tan(C-A)-tan(A-B)tan(B-C)tan(C-A)/1-tan(A-B)tan(B-C)-tan(C-A)tan(A-B)-tan(B-C)tan(C-A) = 0 [ from (1) ]
or, tan(A-B)+tan(B-C)+tan(C-A)-tan(A-B)tan(B-C)tan(C-A) = 0 × 1-tan(A-B)tan(B-C)-tan(C-A)tan(A-B)-tan(B-C)tan(C-A)
or, tan(A-B)+tan(B-C)+tan(C-A)-tan(A-B)tan(B-C)tan(C-A) = 0
or, tan(A-B)+tan(B-C)+tan(C-A) = tan(A-B)tan(B-C)tan(C-A)
LHS = RHS
Hence, proved