Question

tan²A-tan²B=sin²A-sin²B/cos²Acos²B​

Answer 1

Step-by-step explanation:

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

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Student Answers

LOCHANA2500 | STUDENT

tan²A - tan²B ≠ (sin²A - sin²B)/(cos²A- cos²B)

but, tan²A - tan²B = (sin²A - sin²B)/cos²A.cos²B

L:H:S ≡ tan²A - tan²B

⇒ use tan²x = sec²x -1

= sec²A - 1 - (sec²B - 1)

= sec²A - 1 - sec²B +1

= sec²A - sec²B

= 1/cos²A - 1/cos²B

= (cos²B - cos²A)/cos²A.cos²B

⇒ use cos²x =1 - sin²x

= (1-sin²B -1 + sin²A)/cos²A.cos²B

= (sin²A - sin²B)/cos²A.cos²B

Hence L:H:S ≡ R:H:S

Answer 2

answer in the aatachment