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

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(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

Vedclass · Accepted Answer

$tan \,8	heta\, cot\, 2	heta$

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 $\frac{2 \sin^2 4 	heta}{\cos 8 	heta} 	imes \frac{\cos 4 	heta}{2 \sin^2 2 	heta}$.Using $\sin 4 	heta = 2 \sin 2 	heta \cos 2 	heta$,we get $\frac{2 (2 \sin 2 	heta \cos 2 	heta) \sin 4 	heta \cos 4 	heta}{\cos 8 	heta (2 \sin^2 2 	heta)}$.Simplifying,$\frac{2 \sin 4 	heta \cos 4 	heta}{\cos 8 	heta} 	imes \frac{\cos 2 	heta}{\sin 2 	heta} = \frac{\sin 8 	heta}{\cos 8 	heta} 	imes \cot 2 	heta = 	an 8 	heta \cot 2 	heta$.

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

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