1.

Prove that: `(sec4A-1)/(sec8A-1)=tan2A. cot8A`

Answer» LHS= `(sec4A-1)/(sec8A-1)`
`=(1/(cos4A)-1)/(cos8A)=((1-cos4A)/(cos4A))/((1-cos8A)/(cos8A))`
`(1-cos4A)/(1-cos8A).(cos8A)/(cos4A) (therefore 1-cos2theta=2sin^(2)theta)`
`(2sin^(2)2A)/(2sin^(2)4A).(cos8A)/(cos4A)`
`=(2sin^(2)2A.cos8A)/(sin4A.(2sin4Acos4A))`
`=(2sin^(2)2A.cos8A)/((2sin2Acos2A)sin8A)`
`(sin2A)/(cos2A).(cos8A)/(sin8A)`
`=tan2A.cot8A`
=RHS Hence Proved


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