Question

prove that tan^2A - tan^2 B = sin^2 A - sin^2 B / cos^2 B cos^2A

Answer 1

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

Answer 2

$\huge\mathbb{\red A\pink{N}\purple{SW} \blue{ER} \orange{.. }}$

See the attachment please

...................