Prove that: (1/sec^2A - cos^2A + 1/cosec^2A - sin^2A)sin^2Acos^2A = 1 - sin^2Acos^2A / 2 + sin^2Acos^2A

Asked by  | 11th Sep, 2012, 07:37: PM

Expert Answer:

LHS = [1/(sec^2A - cos^2A ) + 1/(cosec^2A - sin^2A ] sin^2A .cos^2A
= [cos^2A/(1-cos^4A) + sin^2A/(1-sin^4A)] * sin^2Acos^2A
= [cos^2A/(1-cos^2A)(1+cos^2A) + sin^2A/ (1-sin^2A)(1+sin^2A)] *sin^2A .cos^2A
= [cos^2A/(sin^2A)(1+cos^2A) + sin^2A/ (cos^2A)(1+sin^2A)] *sin^2A .cos^2A
= (cos^4A/1+cos^2A) + (sin^4A/1+sin^2A)
= (cos^4A+sin^2Acos^4A+sin^4A+cos^2Asin^4A / (2+sin^2Acos^2A)
= cos^4A+sin^4A+cos^2Asin^2A(sin^2A+cos^2A / (2+sin^2Acos^2A)
= cos^4A+sin^4A+cos^2Asin^2A / (2+sin^2Acos^2A)
= [(cos^2A+sin^2A)^2 - sin^2Acos^2A] / (2+sin^2Acos^2A)
= (1-sin^2Acos^2A) / (2+sin^2Acos^2A)
= RHS

Answered by  | 11th Sep, 2012, 10:25: PM

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