Question

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

Answer 1

Step-by-step explanation:

put A,B = 0°

0 -0 = 0- 0/1 , 0 = 0 satisfied

Answer 2

hey mate here is ur answer

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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