Question

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Answer 1

Answer:

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

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Here is your answer......



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