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

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Question

(B) Given expression: $\frac{\sec 8A - 1}{\sec 4A - 1}$$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} 	imes \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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