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Prove that `(Sec8A-1)/(Sec4A-1)=(tan8A)/(tan2A)`

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Prove that `(Sec8A-1)/(Sec4A-1)=(tan8A)/(tan2A)`
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`L.H.S. = (sec8A - 1)/(sec4A - 1) = ((1-cos8A)/(cos8A))/((1-cos4A)/(cos4A))`
`=((2sin^2 4A)cos4A)/((2sin^2 2A)cos8A)`
`=(sin8Asin4A)/(cos8A(2sin^2 2A))`
`=(tan8A(2sin2Acos2A))/(2sin^2 2A)`
`=tan8Acot2A`
`=(tan8A)/(tan2A) = R.H.S.`
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