Question

(1/sec^2A-cos^A +1/cosec^2A-sin^2A) sin^2A×cos^2A​

Answer 1

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Answer Text

LHS

=⎛⎝11cos2A−cos2A+11sin2A−sin2A⎞⎠sin2A.cos2A

=(cos2A1−cos4A+sin2A1−sin4A)sin2Acos2A

=(cos2A(1−cos2A)(1+cos2A)+sin2A(1−sin2A)(1+sin2A))sin2Acos2A

=cos2Asin2Acos2Asin2A(1+cos2A)+sin2Asin2Acos2AcosA(1+sin2A)

=cos4A1+cos2A+sin4A1+sin2A

=cos4A(1+sin2A)+sin4A(1+cos2A)(1+cos2A)(1+sin2A)

=cos4A+sin2A.cos4A+sin4A+sin4Acos2A1+cos2A+sin2A+sin2Acos2A

=(cos2A)2+(sin2A)2+sin2Acos2A(cos2A+sin2A)2+sin2Acos2A

by using identity, (a+b)2=a2+b2+2ab

=(cos2A+sin2A)2−2cos2Asin2A+sin2Acos2A2+sin2Acos2A

=1−sin2Acos2A2+sin2Acos2A

= RHS

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Answer 2

Answer:

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