1.

`(1/(sec^2A-cos^2A)+1/(cosec^2A-sin^2A))sin^2Acos^2A=(1-sin^2Acos^2A)/(2+sin^2Acos^2A)`

Answer» LHS
`=(1/(1/cos^2A -cos^2A) + 1/(1/sin^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^2Acos^2A`
`= (cos^2Asin^2Acos^2A)/(sin^2A(1+cos^2A)) + (sin^2A sin^2Acos^2A)/(cos^A(1+sin^2A))`
`= cos^4A/(1+ cos^2A) + sin^4A/(1+sin^2A)`
`= (cos^4A(1+sin^2A) + sin^4A(1+ cos^2A))/((1+cos^2A)(1+sin^2A))`
`= (cos^4A + sin^2A.cos^4A + sin^4A + sin^4Acos^2A)/(1 + cos^2A + sin^2A + sin^2Acos^2A)`
`= ((cos^2A)^2 + (sin^2A)^2 + sin^2Acos^2A(cos^2A+sin^2A))/(2 + sin^2Acos^2A)`
by using identity, `(a+b)^2 = a^2 + b^2 + 2ab`
`= ((cos^2A + sin^2A)^2 - 2cos^2Asin^2A + sin^2Acos^2A)/(2 + sin^2Acos^2A)`
`= (1- sin^2Acos^2A)/(2+sin^2Acos^2A) `
= RHS


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