give that sin( A + B )= 2cos(A - B ) and tan A = 1/3,Find the exact value of tanB

if tanA = 1/3, then by the Pythagorean triangle

sinA = 1/√10 , and cosA = 3/√10

sin(A+B) = 2cos(A-B)
sinAcosB + cosAsinB = 2cosAcosB + 2sinAsinB
(1/√10)cosB + (3/√10)sinB = 2(3√10)cosB + 2(1/√10)sinB
times √10 ....
cosB + 3sinB = 6cosB + 2sinB
-sinB = 3cosB
-sinB/cosB = 3
tanB = -3

Given:

sin( A + B )= 2cos(A - B )
Expand by sum and difference formulae:
sinAcosB+cosAsinB=2cosAcosB+2sinAsinB
Divide each side by cosAcosB and simplify:
2tanAtanB-tanA-tanB+2=0
Substitute tanA=1/3
(2/3)tanB-tanB=5/3
Solve for tanB
tanB=5

found a silly arithmetic error in 3rd last line, should have been

sinB = 5cosb
sinB/cosB = 5
tanB = 5

To find the exact value of tan(B), we need to use the given information and apply trigonometric identities.

Given:
sin(A + B) = 2cos(A - B) (Eq. 1)
tan(A) = 1/3

We can start by manipulating Eq. 1 to express sin(A + B) and cos(A - B) in terms of sine and cosine of A and B. We'll use the angle addition and subtraction formulas:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B) (Eq. 2)
cos(A - B) = cos(A)cos(B) + sin(A)sin(B) (Eq. 3)

Let's substitute these equations into Eq. 1:

sin(A)cos(B) + cos(A)sin(B) = 2(cos(A)cos(B) + sin(A)sin(B))

Now, let's rearrange the terms to isolate sin(B):

sin(A)cos(B) + cos(A)sin(B) = 2cos(A)cos(B) + 2sin(A)sin(B)

sin(A)cos(B) - 2sin(A)sin(B) = 2cos(A)cos(B) - cos(A)sin(B)

Factor out sin(B) and cos(B):

sin(A)[cos(B) - 2sin(B)] = cos(A)[2cos(B) - sin(B)]

Divide both sides by cos(A)[2cos(B) - sin(B)]:

sin(A) / cos(A) = cos(A) / [2cos(B) - sin(B)]

Using the identity tan(A) = sin(A) / cos(A), we can substitute the value tan(A) = 1/3:

1/3 = cos(A) / [2cos(B) - sin(B)]

Cross-multiply:

3cos(A) = 2cos(B) - sin(B)

Now, we have an equation involving cos(A) and cos(B). We'll use the Pythagorean identity to relate these trigonometric values:

cos^2(A) + sin^2(A) = 1

Since tan(A) = 1/3, we have sin(A) = 1 and cos(A) = 3/sqrt(10).

Substituting these values into the equation:

3(3/sqrt(10)) = 2cos(B) - sin(B)

9/sqrt(10) = 2cos(B) - sin(B) (Eq. 4)

Now, let's solve Eq. 4 for cos(B) - sin(B):

2cos(B) - sin(B) = 9/sqrt(10)

Move the sin(B) term to the left side:

2cos(B) = sin(B) + 9/sqrt(10)

Now, we'll square both sides of the equation to eliminate the square root:

(2cos(B))^2 = (sin(B) + 9/sqrt(10))^2

4cos^2(B) = sin^2(B) + (9/sqrt(10))^2 + 2sin(B)(9/sqrt(10))

Using the Pythagorean identity again, we substitute cos^2(B) = 1 - sin^2(B):

4(1 - sin^2(B)) = sin^2(B) + (9/sqrt(10))^2 + 2sin(B)(9/sqrt(10))

Distribute and rearrange terms:

4 - 4sin^2(B) = sin^2(B) + 81/10 + (18/10)(sin(B))

Multiply through by 10 to remove the fraction:

40 - 40sin^2(B) = 10sin^2(B) + 81 + 18sin(B)

Combine like terms:

50sin^2(B) + 18sin(B) - 40 = 0

This equation is quadratic in form. We can solve it using factoring, the quadratic formula, or graphing techniques. I will use factoring for simplicity.

Factor the quadratic:

(10sin(B) + 20)(5sin(B) - 2) = 0

Setting each factor equal to zero:

10sin(B) + 20 = 0 or 5sin(B) - 2 = 0

First equation:
10sin(B) = -20
sin(B) = -2

Since the sine function is only defined in the range [-1, 1], there are no solutions for sin(B) = -2.

Second equation:
5sin(B) = 2
sin(B) = 2/5

Now, we can use the identity tan(B) = sin(B) / cos(B) to find the exact value of tan(B):

tan(B) = (2/5) / cos(B)

However, with the given information, we don't have enough data to determine the exact value of cos(B) or find a numerical value for tan(B). More information or additional trigonometric relationships would be required to obtain an exact answer.