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Question
If `(cos^4alpha)/(cos^2beta) + (sin^4alpha)/(sin^2beta)` = 1, prove that `(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha)` = 1
Solution
`(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha) = (cos^2beta*cos^2beta)/(cos^2alpha) + (sin^2beta*sin^2beta)/(sin^2alpha)` ......(1)
Given `(cos^4alpha)/(cos^2beta) + (sin^4alpha)/(sin^2beta)` = 1
`(cos^4alpha*sin^2beta + sin^4alpha*cos^2beta)/(cos^2beta sin^2beta)` = 1
`cos^4alpha sin^2beta + sin^4alpha*cos^2beta = cos^2beta sin^2beta`
`cos^4alpha(1 - cos^2beta) + (1 - cos^2alpha)^2 cos^2beta = cos^2beta(1 - cos^2beta)`
`cos^4alpha - cos^4alpha cos^2beta + (1 - 2cos^2alpha + cos^4alpha) cos^2beta = cos^2beta - cos^4beta`
`cos^4alpha - cos^4alpha cos^2beta + cos^2beta - 2cos^2alpha cos^2beta + cos^4alpha cos^2beta = cos^2beta - cos^4beta`
`cos^4alpha - 2cos^2alpha cos^2beta + cos^4beta` = 0
`(cos^2alpha - cos^2beta)^2` = 0
`cos^2alpha = cos^2beta` ......(2)
`1 - sin^2alpha = 1 - sin^2beta`
`sin^2alpha = sin^2beta` ......(3
Using equation (2) and (3), equation (1) becomes
`(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha) = (cos^2beta*cos^2beta)/(cos^2alpha) + (sin^2beta*sin^2beta)/(sin^2alpha)`
= `(cos^2beta*cos^2alpha)/(cos^2alpha) + (sin^2beta*sin^2alpha)/(sin^2alpha)`
= `cos^2beta + sin^2beta`
= 1
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