Prove that `(sec 8theta-1)/(sec 4 theta -1) = (tan 8theta)/(tan 2theta)`

Explanation

LHS`=(sec 8theta-1)/(sec4theta-1)`
`=((1)/(cos 8theta)-1)/((1)/(cos 4theta)-1)=(1-cos 8theta)/(cos 8theta)xx(cos 4theta)/(1-cos 4theta)`
`=(2sin^(2)4theta)/(cos 8theta)xx(cos 4theta)/(2sin^(2)2theta)`
`=(2sin^(2)4theta)/(cos 8theta)xx(cos 4theta)/(2sin^(2)2theta)`
`[because 1-cos8theta=2sin^(2)""(8theta)/(2)=2sin^(2)4theta "and" 1-cos 4 theta=2 sin^(2)""(4theta)/(2)=2sin^(2)2theta]`
`=((2sin 4theta cos 4theta))/(cos 8theta)xx(sin4theta)/(2sin^(2)2theta)`
`=((2sin4thetacos 4theta)/(cos 8 theta))xx((2sin2thetacos 2theta)/(2sin^(2)2theta))`
`=((sin2(4theta))/(cos 8theta))xx(cos 2theta)/(sin 2theta))`
`=((sin 8theta)/(cos 8theta))xx(cos 2 theta)/(sin 2theta))`
`=tan 8theta cot 2 theta=(tan 8theta)/(tan 2theta)`=RHS.