trigonometry
posted by lindsay .
Given that tan A = 2 tan B (B is the symbol for Beta) show that tan (AB) = sin 2B/3 cos2 Beta.

tan(AB) = (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.
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