Question

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

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