Question

Given that tan A = 2 tan B (B is the symbol for Beta) show that tan (A-B) = sin 2B/3- cos2 Beta.

Answer 1

tan(A-B) = (tanA - tanB)/(1+tanAtanB)
but tanA = 2tanB
so above
= (2tanB - tanB)/(1 + 2tanBtanB)
= tanB/(1+2tan^2B)
= (sinB/cosB) / (1 + 2sin^2 B /cos^2 B)
= sinB/cosB /[(cos^2 B + 2sin^2 B)/cos^2 B]

= (sinB/cosB) * (cos^2 B)/(cos^2 B + 2sin^2 B)
= sinBcosB /(cos^2 B + 2(1 - cos^2 B) )
= sinBcosB / (2 - cos^2 B)
= 2sinBcosB/ (4 - 2cos^2 B)
= sin 2B / (4 - 2cos^2 B)

which is different from yours,
I can't seem to find my error if there is one.