sin^2a sec^2b +tan^2b cos^2a= sin^2a+tan^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
NEELA | STUDENT
Kindly note that the identity to be proved is tan^2 A - tan^2 B = (sin^2 A - sin^2 B)/(cos^2 A * cos^2 B) and not tan^2 A - tan^2 B = (sin^2 A - sin^2 B)/(cos^2 A- cos^2 B).
The latter is not an identity as: If A = 60 degree and B =30 degree, LHS =3-1/3 = 2 and 2/3. The RHS = (0.75-0.25)/(0.25-0.75) = -1, which disproves that tan^2 A - tan^2 B = (sin^2 A - sin^2 B)/(cos^2 A- cos^2 B) is an identity.
Also:
LHS=tan^2x-tan^2B = (sinA/cosB)^2-(sinB/cosB)^2
=(sin^2A*cos^2B-sin^2B*cs^2A)/(cos^2A*cos^2B)
=[sin^2A(1-sin^2B)-sin^2B(1-sin^2A)]/(cos^2A*cos^2B)
=[Sin^2A-sin^2B]/{cos^2A*cos^2B} Therefore,
tan^2 A - tan^2 B = (sin^2 A - sin^2 B)/(cos^2 A * cos^2 B) is an identity
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Step-by-step explanation:
sin^2a.sec^2b+tan^2b.cos^2a=sin^2a+tan^2b
