यदि ` (cos^(4)A)/(cos^(2)B) + (sin^(4)A)/(sin^(2)B)= 1`, तो सिद्ध कीजिए कि -
1.` sin^(4) A + sin^(4) B = 2 sin^(2) A sin^(2) B`, " "
2. `(cos^(4)B)/(cos^(2)A) + (sin^(4)B)/(sin^(2)A)=1 `

Explanation

` (cos^(4) A)/(cos^(2)B)+ (sin^(4)A)/(sin^(2) B) = 1 `
` rArr ( cos^(4) A sin^(2) B + sin^(4) A cos ^(2) B)/( cos^(2) B sin^(2) B)=1`
` rArr ( 1-sin^(2) A )^(2) sin^(2) b + sin^(4) A ( 1- sin^(2) B )`
` = sin^(2) B ( 1- sin^(2) B)`
` rArr ( 1 + sin^(4) A - 2 sin^(2 ) A ) sin^(2) B + sin^(4) A `
` - sin^(4) A sin^(2) B = sin^(2) B - sin^(4) B`
` rArr sin^(2) B + sin^(4) A sin^(2) B - 2 sin ^(2) A sin^(2) B `
` + sin^(4) A - sin^(4) A sin^(2) B = sin^(2) B = sin^(2) B - sin^(4) B`
` rArr sin^(4) A + sin^(4) B = 2 sin^(2) A sin^(2) B` जो पहला भाग है ।
`rArr sin^(4) A + sin^(4) B - 2 sin ^(2) A sin^(2) B = 0`
`rArr ( sin^(2) A - sin^(B) )^(2) = 0 `
`rArr sin^(2) A - sin^(2) B = 0 `
` rArr sin^(2) A = sin^(2) B ` ... (1)
अब ` L.H.S = ( cos^(4)B)/(cos^(2)A)+ (sin^(4)B)/(sin^(2)A)`
` = ( cos^(2) B* cos^(2) B)/(cos^(2)A) + (sin^(2) B* sin^(2)B)/(sin^(2)A) `
` = (cos^(2) B ( 1 - sin^(2) B))/(1-sin^(2)A) + (sin^(2) B* sin^(2) A )/(sin^(2)A )`
[ समीकरण (1 ) से ]
` (cos^(2) B(1-sin^(2)A))/((1-sin^2)A) = sin^(2) B `
[ समीकरण (1 ) से ]
` cos^(2) B + sin^(2) B`
` 1 = R.H.S `