Question

show and explain the derivation for tan(a+b)

Answer 1

tan(a+b)=(tanA+tanB)/(1-TanA.TanB)

Answer 2

(1-TanA.TanB)?

Answer 3

tan(A+B)



sin(A+B)/cos(A+B)



(sin(A)cos(B)+cos(A)sin(B))/(cos(A)cos(B)-sin(A)sin(B))



((sin(A)cos(B))/(cos(A)cos(B))+(cos(A)sin(B))/(cos(A)cos(B)))/((cos(A)cos(B))/(cos(A)cos(B))-(sin(A)sin(B))/(cos(A)cos(B)))



(sin(A)/cos(A)+sin(B)/cos(B))/(1-(sin(A)/cos(A))(sin(B)/cos(B)))



(tan(A)+tan(B))/(1-tan(A)*tan(B))



So tan(A+B) = (tan(A)+tan(B))/(1-tan(A)*tan(B))