Question

Prove identity of tan(A+B).

Answer 1

Hey mate,..

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//

Hope it will help you.

Answer 2

tan(A+B) = (tanA + tanB) / (1 - tanA.tanB)

T2T :)