"simple" trig problem

[romsofia]

1. The problem statement, all variables and given/known data 2(sinA+cosB)sinB=3-cosB. Find 3tan2A+4tan2B




2. Relevant equations Trig Identities



3. The attempt at a solution well, i expanded 3tan2A+4tan2B= 3sin2Acos2B+4sin2Bcos2A all over cos2Acos2B after that I am stuck.

[n.karthick]

I didn't get you. How you expanded tan^2 A?
How will u express tan A in terms of sin A and cos A?

[romsofia]

I didn't get you. How you expanded tan^2 A?
How will u express tan A in terms of sin A and cos A?

Ok, 3tan2A= 3sin2A/cos2A and 4tan2B= 4sin2B/cos2B, so 3sin2A/cos2A+4sin2B/cos2B= 3sin2Acos2B+4sin2Bcos2A all over cos2Acos2B

[The Chaz]

... 2(sinA+cosB)sinB=3-cosB. Find 3tan2A+4tan2B

...

Are you sure that one of the cosines isn't "cosA" ???