If `(cos^4A)/(cos^2B)+(sin^4A)/(sin^2B)=1`, then prove that
(i) `sin^4A+sin^4B=2 sin^2Asin^2B`
(ii) `(cos^4B)/(cos^2A)+(sin^4B)/(sin^2A)=1`

Explanation

Given,
`(cos^A)/(cos^2B)+(sin^4A)/(sin^2B)=1=(cos^2A+sin^2A)`
or `(cos^A)/(cos^2B)-1=cos^2A=sin^2A-(sin^4A)/(sin^2B)`
or `(cos^2A(cos^2A-cos^2B))/cos^2B=sin^2A((sin^2B-sin^2A))/sin^2B`
or `(cos^2A)/(cos^2B)(cos^2A-cos^2B)=sin^2A/sin^2B[(1-cos^2B)-(1-cos^2A)]`
or `cos^2A/cos^2B(cos^2A-cos^2B)=sin^2A/sin^2B(cos^2A-cos^2B)`
or `(cos^2A-cos^2B)(cos^2A/cos^2B-sin^2A/sin^2B)=0`
Whem `cos^2A-cos^2B=0`, we have
`cos^2A=cos^2B (i)`
When `cos^2A/(cos^2B-sin^2A/sin^2B=0`,we have
`cos^2Asin^2B=sin^2A cos^2B`
or `cos^2A(1-cos^2B)=(1-cos^2A)cos^2B`
or `cos^2A-cos^2Acos^2B=cos^2B-cos^2Acos^2B`
? or `cos^2A=cos^2B (ii)`
Thus, in both the cases, `cos^2A=cos^2B`. Therefore,
`:. 1-sin^2A=1-sin^2B" or " sin^2A =sin^2B (iii)`
(i) `L.H.S.=sin^4A+ sin^4B`
`=(sin^2A-sin^2B)^2+2sin^2Asin^2B`
`=2sin^2Asin^2B=R.H.S. [ :'sin^2A=sin^2B]`
(ii) L.H.S. `=cos^4B/cos^2A+sin^4B/sin^2A=cos^4B/cos^2B+sin^4B/sin^2B`
`=cos^2B+sin^2B=1=R.H.S.`