Question

3-4sin2%cos2=3-tan2
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Answer 1

Answer:

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

Hence Proved

Hope this helps...:)

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