If `(cos^4alpha)/(cos^2beta)+(s in^4alpha)/(s in^2beta)=1` prove that `s in^4alpha+s in^4beta=2s in^2alphas in^2beta` `(cos^4beta)/(cos^2alpha)+(s in^4beta)/(s in^2alpha)=1`

Explanation

In such type of questions
Remember that
[1 can be replaced by `(sin^(2)alpha+cos^(2)alpha)`]
Using this
`(cos^(4)alpha)/(cos^(2)beta)+(sin^(4)alpha)/(sin^(2)beta)=cos^(2)alpha+sin^(2)alpha`
`(cos^(2)alpha)/(cos^(2)beta)-cos^(2)alpha+(sin^(4)alpha)/(sin^(2)beta)-sin^(2)alpha=0`
`cos^(2)alpha[(cos^(2)alpha-cos^(2)beta)/(cos^(2)beta)]-sin^(2)alpha[(cos^(2)alpha-cos^(2)beta)/(sin^(2)beta)]=0`
`(cos^(2)alpha-cos^(2)beta)[(cos^(2)alpha-cos^(2)beta)/(sin^(2)beta)]=0`
From second factor
`cos^(2)alpha(1-cos^(2)beta)-cos^(2)beta(1-cos^(2)alpha)=0`
`cos^(2)alpha-cos^(2)beta=0`
Now using this result,
`(sin^(4)beta)/(sin^(2)alpha)+(cos^(4)beta)/(cos^(2)alpha) =(sin^(4)alpha)/(sin^(2)alpha) + (cos^(4)alpha)/(cos^(2)alpha) (therefore sin^(2)alpha=sin^(2)beta, cos^(2)alpha=cos^(2)beta)`
`=sin^(2)alpha+cos^(2)alpha`
=1