Prove that (3 - 4sin^2 A)/cos^2A= 3-tan^2A
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Let theta be 'A'
Given=> 3 - 4sin²A/cos²A = 3 - tan²A
LHS
=> (3 - 4sin²A)/cos²A
=> [3 - 4(1 - cos²A)]/cos²A
=> (3 - 4 + 4cos²A)/cos²A
=> (4cos²A - 1)/cos²A
=> [3cos²A + (cos²A - 1)]/cos²A
=> (3cos²A - sin²A)/cos²A
=> 3cos²A/cos²A - sin²A/cos²A
=> 3 - tan²A
=> RHS
. '. LHS = RHS.
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