Question

cos^2a-cos^2b/cos^2a.cos^2b=tan^2b-tan^2​a

Answer 1

Answer:

your qsn is wrong-- right qsn is

(sin^2A - sin^B)/cos^A.cos^B=tan^2A - tan^B

here is your answer----------

Step-by-step explanation:

LHS = tan²A - tan²B

={ sin²A/cos²A } - { sin²B/cos²B }

= {sin²A.cos²B - sin²B.cos²A }/cos²A.cos²B

we know,

sin²x + cos²x = 1

so,

cos²B = 1 - sin²B

cos²A = 1 - sin²A

use this here,

= {sin²A (1 - sin²B) - sin²B(1 - sin²A)}/cos²A.cos²B

= { sin²A - sin²A.sin²B - sin²B + sin²A.sin²B }/cos²A.cos²B

= ( sin²A - sin²B )/cos²A.cos²B = RHS

hope it wil help you