If 2tana=3tanb, then prove that tan(a-b)=sin2b/(5-cos2b)

Tan ( A - B ) = ( tan A - tan B ) / ( 1 + tan A tan B )
tan A = 3 tan B/2
tan ( A - B ) = ((3 tan B/ 2)-tan B) / ( 1 + 3 tan² B/2)=
= (tan B/2)  / ( 2 + 3 sin²B/cos²B )=
= (sin B / cos B) / (( 2cos² B+3sin²B)/cos²B)=
=( sin B cos B ) / ( 2 cos²B + 3 ( 1 - cos² B ) ) =
= (sin B cos B ) / ( 2 cos² B + 3 - 3 cos² B ) =
= ( sin 2 B ) / 2 ( 3 - cos² B ) =
= ( sin 2 B ) / ( 6 - cos² B )=
= ( sin 2 B ) / ( 5 + 1 - 2 cos² B )=
= ( sin 2 B ) / ( 5 + sin² B + cos ² B - 2 cos² B ) =
= ( sin 2 B ) / ( 5 - ( cos² B - sin² B ) ) =
= ( sin 2 B ) / ( 5 - cos 2 B )  - correct 

It says in the questions that it is 6.5?