prove that sec 8A-1/sec 4A-1=tan 8a/tan 2A

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Solution

L.H.S. =
(sec 8A -1) / (sec 4A -1)
$= \frac{(\frac{1}{\cos 8 A} - 1)}{(\frac{1}{\cos 4 A} - 1)}$
$= \frac{\frac{1 - \cos 8 A}{\cos 8 A}}{\frac{1 - \cos 4 A}{\cos 4 A}} = (\frac{2 \sin^{2} 4 A}{2 \sin^{2} 2 A}) (\frac{\cos 4 A}{\cos 8 A}) = (\frac{2 \sin 4 A . \cos 4 A \times \sin 4 A}{\cos 8 A}) (\frac{1}{2 \sin^{2} 2 A}) = (\frac{\sin 8 A}{\cos 8 A} . \sin 4 A) (\frac{1}{2 \sin^{2} 2 A}) = \tan 8 A . (\frac{2 \sin 2 Acos 2 A}{2 \sin^{2} 2 A}) = \tan 8 A (\frac{\cos 2 A}{\sin 2 A}) = \frac{\tan 8 A}{\tan 2 A} = RHS$

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