Question

prove that: (1/sec2a-cos2a+1/cosec2a-sin2a)sin2acos2a​

Answer 1

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)

cross multiplying

= (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

3

Step-by-step explanation:

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Answer 2

Answer:

Your answer attached in the photo

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