Question

Show that tan² A-tan² B = sin² A-sin²B/cos²A•cos²B​

Answer 1

Answer:

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

Answer 2

hope this answer will be helpful ♥️♥️♥️