Math, asked by brainlyuser9765, 10 months ago

prove that, tan^2 A+ tan^2 B = sin(A+B). sin (A-B)/ cos^2 A .cos ^2B​

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Answered by Anonymous
2

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

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