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

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(B) Given expression: $\frac{\sec 8A - 1}{\sec 4A - 1}$Using $\sec 	heta = \frac{1}{\cos 	heta}$,we get:$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\co

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

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(B) Given expression: $\frac{\sec 8A - 1}{\sec 4A - 1}$Using $\sec 	heta = \frac{1}{\cos 	heta}$,we get:$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} \cdot \frac{\cos 4A}{1 - \cos 4A}$Using the identity $1 - \cos 2	heta = 2\sin^2 	heta$:$= \frac{2\sin^2 4A}{\cos 8A} \cdot \frac{\cos 4A}{2\sin^2 2A}$$= \frac{(2\sin 4A \cos 4A) \cdot \sin 4A}{\cos 8A \cdot 2\sin^2 2A}$Using $\sin 2	heta = 2\sin 	heta \cos 	heta$,so $\sin 8A = 2\sin 4A \cos 4A$:$= \frac{\sin 8A}{\cos 8A} \cdot \frac{\sin 4A}{2\sin^2 2A}$$= 	an 8A \cdot \frac{2\sin 2A \cos 2A}{2\sin^2 2A}$$= 	an 8A \cdot \frac{\cos 2A}{\sin 2A} = 	an 8A \cdot \cot 2A = \frac{	an 8A}{	an 2A}$

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

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