frac{{sec 8A - 1}}{{sec 4A - 1}} = | vedclass

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Question

(b) $\frac{{\sec 8A - 1}}{{\sec 4A - 1}}$

$ = \frac{{1 - \cos 8A}}{{\cos 8A}}.\frac{{\cos 4A}}{{1 - \cos 4A}}$

$ = \frac{{2{{\sin }^2}4A}}{

Vedclass · Accepted Answer

$\frac{{	an 8A}}{{	an 2A}}$

Vedclass · Answer

(b) $\frac{{\sec 8A - 1}}{{\sec 4A - 1}}$

$ = \frac{{1 - \cos 8A}}{{\cos 8A}}.\frac{{\cos 4A}}{{1 - \cos 4A}}$

$ = \frac{{2{{\sin }^2}4A}}{{\cos 8A}}\frac{{\cos 4A}}{{2{{\sin }^2}2A}}$

$ = \frac{{2\sin 4A\cos 4A}}{{\cos 8A}}\frac{{\sin 4A}}{{2{{\sin }^2}2A}}$

$ = 	an 8A\frac{{2\sin 2A\cos 2A}}{{2{{\sin }^2}2A}} $

$= \frac{{	an 8A}}{{	an 2A}}.$

$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $

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