Find 3tan^2A+4tan^2B for A,B Real Numbers

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SUMMARY

The discussion focuses on solving the expression 3tan²A + 4tan²B, given the equation 2(sinA + cosB)sinB = 3 - cosB. Participants clarify that A and B are real numbers, and the solution involves expanding the expression using trigonometric identities. The key transformation used is 3tan²A + 4tan²B = 3sin²A*cos²B + 4sin²B*cos²A, which is derived by manipulating the terms over cos²A*cos²B. The discussion emphasizes the importance of correctly interpreting variable notation in trigonometric contexts.

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romsofia
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1. Homework Statement A and B are real numbers satisfying 2(sinA+cosB)sinB=3-cosB. Find 3tan^2A+4tan^2B
2. Homework Equations Trig Identities
3. The Attempt at a Solution well, i expanded 3tan^2A+4tan^2B= 3sin^2A*cos^2B+4sin^2B*cos^2A all over cos^2A*cos^2B after that I am stuck :x
 
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Are we to assume that "A" is the same as "a" and that "B" is the same as "b"?
 
HallsofIvy said:
Are we to assume that "A" is the same as "a" and that "B" is the same as "b"?

yes, let me edit that :x
 

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