Question

3-4 sine
cose
3-tan- .​

Answer 1

Answer:

$\huge\boxed{3 - tan²A}$

Step-by-step explanation:

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

Hence Proved

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