Question

tan²A-tan²B=cos²B-cos²A/cos²B cos²A=sin²A-sin²B/cos²B cos²A,Prove it by using trigonometric identities.

Answer 1

Given :

tan²A - tan²B = cos²B - cos²A/cos²B cos²A= sin²A- sin²B / cos²B cos²A

LHS = tan²A - tan²B

= sin²A/cos²A - sin²B/cos²B

[tanθ = sinθ /cosθ ]

= (sin²A cos²B - sin²B cos²A)/cos²A cos²B

=[ sin²A(1-sin²B) - sin²B(1-sin²A)]/cos²A cos²B

[cos²θ = 1 - sin²θ ]

=[ (sin²A - sin²Asin²B) - (sin²B - sin²Asin²B)]/cos²A cos²B

= [(sin²A - sin²Asin²B - sin²B + sin²Asin²B)]/cos²A cos²B

=[ (sin²A - sin²B - sin²Asin²B + sin²Asin²B)] /cos²A cos²B

LHS = (sin²A - sin²B)/cos²A cos²B = RHS

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Hi ,

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We know that ,

1 ) tanA = sinA/cosA

2 ) sin²A + cos²A = 1

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tan²A - tan²B

= ( sin²A/cos²A ) - ( sin²B/cos²B )

= [(sin²Acos²B - sin²Bcos²A)/cos²Acos²B ]

= [(1-cos²A)cos²B-(1-cos²B)cos²A]/cos²Acos²B

=[cos²B-cos²Acos²B-cos²A+cos²Acos²B]/cos²Acos²B

= (cos²B - cos²A)/cos²Acos²B --( 1 )

=[(1 - sin²B)-(1-sin²A)]/[cos²Bcos²A ]

= ( 1 - sin²B - 1 + sin²A )/(cos²Bcos²A )

= ( sin²A - sin²B )/cos²Bcos²A----( 2 )

I hope this helps you.

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