सिद्ध कीजिए कि - `(sec 8theta - 1)/(sec 4theta -1)=(tan 8theta)/(tan 2theta)`.

Explanation

`L.H.S. = (sec 8theta -1)/(sec 4theta -1)=(1-cos 8theta)/(1-cos 4theta)xx(cos 4theta)/(cos 8theta)`
`=(2sin^(2)4theta)/(2sin^(2)2theta).(cos 4theta)/(cos 8theta)" " [because 1-cos 2theta = 2sin^(2)theta]`
`=((2sin 4theta.cos 4theta).sin 4theta)/(cos 8theta.2sin^(2)2theta)`
`= (sin 8theta)/(cos 8theta).(2 sin 2theta cos 2theta)/(2sin^(2)2theta) " " [because 2 sin theta cos theta = sin 2 theta]`
`= (sin 8theta)/(cos 8theta).(cos 2theta)/(sin 2theta)`
`= tan 8theta.cot2theta " " [because (sin theta)/(cos theta)=tan theta` एंव `(cos theta)/(sin theta)=cot theta]`
`=(tan 8theta)/(tan 2theta)" " [because cot theta =(1)/(tan theta)]`
`rArr " " L.H.S. = R.H.S. " "` इति सिद्धम्