how to prove: (sec4A-1)/(sec8A-1)=tan2A/tanA

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Given (sec 4A -1) / (sec 8A -1)=tan 2A/tanA is false
If your question is
(sec 8A -1) / (sec 4A -1) = tan 8A / tan 2A
L.H.S. = (sec 8A -1) / (sec 4A -1)
[(1 - cos 8A)/ cos 8A] / [(1 - cos 4A)/ cos 4A]
[2 sin² 4A / 2 sin² 2A] * [cos 4A / cos 8A]
[(2 sin 4A * cos 4A) * sin 4A / cos 8A] / [2 sin² 2A]
[(sin 8A / cos 8A) * sin 4A] / [2 sin² 2A]
tan 8A * (sin 4A / 2 sin² 2A)
tan 8A * (2sin 2A * cos 2A / 2 sin² 2A)
tan 8A * ( cos 2A / sin 2A)
tan 8A * cot 2A
tan 8A / tan 2A
R.H.S

Trigonometry Functions - http://math.tutorcircle.com/trigonometry/#trigonometric-functions

by Level 8 User (30.1k points)

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