If `(sin^4theta)/a+(cos^4theta)/b=1/(a+b)` , prove that `(sin^8theta)/(a^3)+(cos^4theta)/(b^3)=1/((a+b)^3)` `(sin^(4n)theta)/(a^(2n-1))+(cos^(4n)theta)/(b^(2n-1))=1/((a+b)^(2n-1)),n in N`

Explanation

Given,`(sin^4theta)/a+(cos^4theta)/b=1/(a+b)`


`=>(sin^4theta)/a+((1-sin^2theta)^2)/b=1/(a+b)`

`=>(bsin^4alpha+a(1+sin^4alpha−2sin^2alpha))=1/(a+b)`

`=>bsin^4alpha+a+asin^4alpha−2asin^2alpha=1/(a+b)`

`=>(a+b)sin^4alpha−2asin^2alpha=1/(a+b)−a`

`=>(a+b)sin^4alpha−2asin^2alpha=(1−ab−a^2)/(a+b)`

`=>sin^2alpha=a/(a+b),cos^2alpha=b/(a+b)`

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