Question

Proove sec 8x-1/sec 4x-1=tan8x/tan2x

Answer 1

Trigonometry... just apply rules for expansion of cos 2A  and sin2A in terms of sinA and cosA,,,

$\frac{sec8x-1}{sec4x-1}=\frac{(1-cos8x)cos4x}{cos8x(1-cos4x)}\\\\=\frac{2sin^24x\ cos4x}{cos8x\ *\ 2\ sin^2 2x}=\frac{2sin4x\ cos4x\ *sin4x}{cos8x\ *sin^22x}\\\\=\frac{sin8x*2sin2x\ cos2x}{cos8x\ * sin^22x}=\frac{tan8x}{tan2x}$

Answer 2

Given,

$\frac{\sec8\alpha\;-\;1}{\sec4\alpha\;-\;1}\\\;\\=\frac{\frac{1}{\cos8\alpha}-1}{\frac{1}{\cos4\alpha}-1}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1-\cos8\alpha}{1-\cos4\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1- 1 +2\sin^{2}4\alpha}{1- 1 +2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{2\sin^{2}4\alpha}{2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{\sin^{2}4\alpha}{\sin^{2}2\alpha}\\\;\\=\frac{2\sin4\alpha.\cos4\alpha.2\sin2\alpha.\cos2\alpha}{2\cos8\alpha.\sin^{2}2\alpha}\\\;\\=\frac{\sin8\alpha.\cos2\alpha}{\cos8\alpha.sin2\alpha}\\\;\\=\frac{tan8\alpha}{tan2\alpha}\\\;\\Note:1. cos2\theta= 1-2\sin^{2}\theta\\\;\\\sin2A=2\sin A.\cos A$