Solving "2(sinA+cosB)sinB=3-cosB" Trig Problem

[romsofia]

"simple" trig problem

Homework Statement

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

Homework Equations

Trig Identities

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" ?