frac{sec 8theta - 1}{sec 4theta - 1} is equal | vedclass

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Question

(D) We have the expression $\frac{\sec 8	heta - 1}{\sec 4	heta - 1}$.Converting to cosine,we get $\frac{\frac{1}{\cos 8	heta} - 1}{\frac{1}{\cos 4\

Vedclass · Accepted Answer

$	an 8	heta \cot 2	heta$

Vedclass · Answer

(D) We have the expression $\frac{\sec 8	heta - 1}{\sec 4	heta - 1}$.Converting to cosine,we get $\frac{\frac{1}{\cos 8	heta} - 1}{\frac{1}{\cos 4	heta} - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} 	imes \frac{\cos 4	heta}{1 - \cos 4	heta}$.Using the identity $1 - \cos 2A = 2 \sin^2 A$,we have $1 - \cos 8	heta = 2 \sin^2 4	heta$ and $1 - \cos 4	heta = 2 \sin^2 2	heta$.Substituting these,we get $\frac{2 \sin^2 4	heta}{\cos 8	heta} 	imes \frac{\cos 4	heta}{2 \sin^2 2	heta}$.Expanding $\sin 4	heta = 2 \sin 2	heta \cos 2	heta$,we get $\frac{2(2 \sin 2	heta \cos 2	heta) \sin 4	heta \cos 4	heta}{\cos 8	heta \cdot 2 \sin^2 2	heta} = \frac{2 \sin 4	heta \cos 4	heta}{\cos 8	heta} 	imes \frac{2 \sin 2	heta \cos 2	heta}{2 \sin^2 2	heta}$.This simplifies to $\frac{\sin 8	heta}{\cos 8	heta} 	imes \frac{\cos 2	heta}{\sin 2	heta} = 	an 8	heta \cot 2	heta$.

$\frac{\sec 8\theta - 1}{\sec 4\theta - 1}$ is equal to

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