Question

can anyone please derive tan(A+B) identity

Answer 1

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

Answer 2

meybe...

Answer 3

We know the identities
(i) sin(A+B) = sinA*cosB+cosA*sinB ---- (1)
(ii) cos(A+B) = cosA*cosB-sinA*sinB ---- (2)
sin(A+B)
Therefore tan(A+B) = --------------
cos(A+B)

Substituting from (1) and (2) we get

sinA*cosB+cosA*sinB
tan (A+B) = --------------------------
cosA*cosB-sinA*sinB

Answer 4

Now, we divide both numerator and denominator in right by cosA*cosb and get:

sinA*cosB+cosA*sinB
-------------------
cosA * cosB
tan (A+B) = --------------------------
cosA*cosB-sinA*sinB
------------------
cosA * cosB


sinAcosB cosAsinB
--------- + --------
cosA*cosB cosAcosB
tan(A+B) = --------------------------
cosA*cosB sinA*sinB
--------- - --------
cosA*cosB cosAcosB

Answer 5

this gives us
sinA sinB
----- + ----
cosA cosB
tan(A+B) = ------------------
sinA*sinB
1 - --------
cosAcosB

tanA + tanB
tan(A+B) = ------------------
1 - tanA*tanB


this is the complete derivation....