Question

prove that, tan(B-C) + tan(C - A) + tan(A - B)
= tan(B-C) tan(C-A)tan(A - B)​

Answer 1

Answer:

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//