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

Answer 1

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

Answer 2

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

Answer 3

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

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