Question

prove that 3-4sin^2theta/cos^2theta =3-tan^2theta.
answer in full step
please it's argent

Answer 1

Hola there,

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...:)

Answer 2

Answer:

3 - 4 sin^2 (A)

= 3 - 4 (1 - cos^2 (A))

= 3 - 4 + 4 cos^2 (A)

= 4 cos^2 (A) - 1

= 3 cos^2 (A) + cos^2 (A) - 1

= 3 cos^2 (A) - (1 - cos^2 (A))

= 3 cos^2 (A) - sin^2 (A)

= cos^2 (A) [3 - tan^2 (A)] [Since, sin^2 (A) / cos^2 (A) = tan^2 (A)]

Hence, 3 - 4 sin^2 (A) = cos^2 (A) [3 - tan^2 (A)]

Divide both sides by cos^2 (A).

(3 - 4 sin^2 (A)) ÷ (cos^2 (A)) = 3 - tan^2A [Proved]tep-by-step explanation: