Given that Sin(A+B)=2Cos(A-B) and tan A = 1/3 ,
Find the exact value of tan B.
To find the exact value of tan B, we will need to use the given equation Sin(A+B) = 2Cos(A-B) and the information that tan A = 1/3.
Let's start by using the identities:
Sin(A+B) = Sin(A)Cos(B) + Cos(A)Sin(B)
Cos(A-B) = Cos(A)Cos(B) + Sin(A)Sin(B)
Using the given equation Sin(A+B) = 2Cos(A-B), we can equate the corresponding parts:
Sin(A)Cos(B) + Cos(A)Sin(B) = 2(Cos(A)Cos(B) + Sin(A)Sin(B))
Now, let's substitute the known value for tan A:
Sin(A)/(Cos(A)) * Cos(B) + Cos(A) * Sin(B) = 2(Cos(A)Cos(B) + Sin(A)Sin(B))
Since tan A = Sin(A)/Cos(A), we can substitute 1/3 for Sin(A)/Cos(A):
(1/3) * Cos(B) + Cos(A) * Sin(B) = 2(Cos(A)Cos(B) + Sin(A)Sin(B))
Since tan A = 1/3, we can also conclude that Sin(A) = 1 and Cos(A) = 3. Substituting these values, we get:
(1/3) * Cos(B) + (3) * Sin(B) = 2(3Cos(B) + Sin(B))
Now, let's simplify the equation:
Cos(B)/3 + 3Sin(B) = 6Cos(B) + 2Sin(B)
To solve this equation, we'll collect the terms involving Cos(B) and Sin(B) on one side and the constant terms on the other side:
-5Cos(B) = Sin(B)/3
Taking the ratio of both sides by Cos(B), we have:
-5 = Sin(B)/(3Cos(B))
Since tan B = Sin(B)/Cos(B), we can substitute tan B for Sin(B)/Cos(B):
-5 = tan B/3
Now, let's solve for tan B:
tan B = -5 * 3
tan B = -15
Therefore, the exact value of tan B is -15.