Question

If `(sin^4theta)/a+(cos^4theta)/b=1/(a+b)` Prove that `(sin^8theta)/a^3+(cos^8theta)/b^3=1/(a+b)^3`

Answer 1

`(sin^(4)A)/(a^(3)) + (cos^(8)A)/(b^(3))= 1/(a+b)`
`rArr (a+b).((sin^(4)A)/(a) + (cos^(4)A)/(b))`
`=1=1^(2)=(sin^(2)A+cos^(4)A)`
`rArr sin^(4)A+a/bcos^(4)A+b/asin^(4)A+cos^(4)A=sin^(4)A + cos^(4)A+2sin^(2)A cos^(2)A`
`rArr A/b cos^(4)A-2sin^(2)A cos^(2)A+b/asin^(4)A=0`
`rArr (sqrt(a/b)cos^(2)A-sqrt(b/a)sin^(4)A)=0`
`rArr (sqrt(a/b)cos^(2)A- sqrt(b/a)sin^(2)A)^(2)=0`
`rArr sqrt(a/b)cos^(2)A-sqrt(b/a)sin^(2)A=0`
`rArr sqrt(a/b)cos^(2)A=sqrt(b/a)sin^(2)A`
`rArr a/b=(sin^(2)A)/(cos^(2)A)`
`rArr (sin^(2)A)/(a) = (cos^(2)A)/(b)= (sin^(2)A+cos^(2)A)/(a+b)`
`rArr sin^(2)A=a/(a+b)` and `cos^(2)A=b/(a+b)`
Now, LHS `=(sin^(8)A)/(a^(3))+(cos^(8)A)/(b^(3))`
`=1/a^(3).(a/(a+b))^(4) + 1/b^(3).(b/(a+b))^(4)`
`rArr sin^(2)=a/(a+b)` and `cos^(2)A=b/(a+b)`
Now, LHS `=(sin^(8)A)/(a^(3))+(cos^(8)A)/(b^(3))`
`=1/a^(3). (a/(a+b))^(4)+1/b^(3).(b/(a+b))^(4)`
`=(a/(a+b)^(4))+b/(a+b)^(4)=(a+b)/(a+b)^(4)`
`=1/(a+b)^(3)`= RHS. Hence poved.