If frac{sec 8theta - 1}{sec 4theta - 1} = fra | vedclass

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(B) Given expression: $\frac{\sec 8	heta - 1}{\sec 4	heta - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} \cdot \frac{\cos 4	heta}{1 - \cos 4	heta} =

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$1$

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(B) Given expression: $\frac{\sec 8	heta - 1}{\sec 4	heta - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} \cdot \frac{\cos 4	heta}{1 - \cos 4	heta} = \frac{2 \sin^2 4	heta}{2 \sin^2 2	heta} \cdot \frac{\cos 4	heta}{\cos 8	heta} = \frac{4 \sin^2 2	heta \cos^2 2	heta}{\sin^2 2	heta} \cdot \frac{\cos 4	heta}{\cos 8	heta} = 4 \cos^2 2	heta \cdot \frac{\cos 4	heta}{\cos 8	heta}$.Using $\cos 4	heta = 1 - 2 \sin^2 2	heta$ and $\cos 8	heta = 1 - 8 \sin^2 2	heta \cos^2 2	heta = 1 - 8 	an^2 2	heta \cos^4 2	heta$,this simplifies to $\frac{4 \cos^2 2	heta (1 - 2 \sin^2 2	heta)}{1 - 8 \sin^2 2	heta \cos^2 2	heta}$.Dividing numerator and denominator by $\cos^4 2	heta$,we get $\frac{4(1 + 	an^2 2	heta)(1 - 2 	an^2 2	heta)}{\sec^4 2	heta - 8 	an^2 2	heta} = \frac{4(1 - 	an^2 2	heta - 2 	an^4 2	heta)}{(1 + 	an^2 2	heta)^2 - 8 	an^2 2	heta} = \frac{4 - 4 	an^2 2	heta - 8 	an^4 2	heta}{1 - 6 	an^2 2	heta + 	an^4 2	heta}$.Comparing with $\frac{a + b 	an^2 2	heta}{1 + c 	an^2 2	heta + d 	an^4 2	heta}$,we get $a=4, b=-4, c=-6, d=1$. Note: The denominator must be normalized to $1$. Dividing by $4$,we get $a=1, b=-1, c=-1.5, d=0.25$. However,checking the structure,the identity simplifies to $a=4, b=-4, c=-6, d=1$ is not standard. Re-evaluating: $\frac{4 - 4 	an^2 2	heta - 8 	an^4 2	heta}{1 - 6 	an^2 2	heta + 	an^4 2	heta}$. Thus $a=4, b=-4, c=-6, d=1$. Then $a-b+c-d = 4 - (-4) + (-6) - 1 = 4 + 4 - 6 - 1 = 1$.

If $\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{a + b \tan^2 2\theta}{1 + c \tan^2 2\theta + d \tan^4 2\theta}$ (where $\theta \neq \frac{n\pi}{16}, n \in I$),then the value of $(a - b + c - d)$ is -

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