frac{{sec ,8theta - 1}}{{sec ,4theta&nb | vedclass

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Question

$\frac{\sec 8 	heta-1}{\sec 4 	heta-1}=\frac{\frac{1}{\cos 8 	heta}-1}{\frac{1}{\cos 4 	heta}-1}=\frac{1-\cos 8 	heta}{\cos 8 	heta} 	imes \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{\sec 8 	heta-1}{\sec 4 	heta-1}=\frac{\frac{1}{\cos 8 	heta}-1}{\frac{1}{\cos 4 	heta}-1}=\frac{1-\cos 8 	heta}{\cos 8 	heta} 	imes \frac{\cos 4 	heta}{1-\cos 4 	heta}$

$=\frac{2 \sin ^{2} 4 	heta}{\cos 8 	heta} 	imes \frac{\cos 4 	heta}{2 \sin ^{2} 2 	heta} \quad\left[\because 1-\cos 8 	heta=2 \sin ^{2} \frac{8 	heta}{2}=2 \sin ^{2} 4 	hetaight]$ and

$\left[\because 1-\cos 4 	heta=2 \sin ^{2} \frac{4 	heta}{2}=2 \sin ^{2} 2 	hetaight]$

$=\frac{(2 \sin 4 	heta \cos 4 	heta)}{\cos 8 	heta} 	imes \frac{\sin 4 	heta}{2 \sin ^{2} 2 	heta}$

$=\left(\frac{2 \sin 4 	heta \cos 4 	heta}{\cos 8 	heta}ight) 	imes\left(\frac{2 \sin 2 	heta \cos 2 	heta}{2 \sin ^{2} 2 	heta}ight)$

$=\left(\frac{\sin 2(4 	heta)}{\cos 8 	heta}ight) 	imes\left(\frac{\cos 2 	heta}{\sin 2 	heta}ight)=\left(\frac{\sin 8 	heta}{\cos 8 	heta}ight) 	imes\left(\frac{\cos 2 	heta}{\sin 2 	heta}ight)$

$	an 8 	heta \cot 2 	heta$

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

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