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

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(B) Given expression: $E = \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	het

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$1$

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(B) Given expression: $E = \frac{\sec 8	heta - 1}{\sec 4	heta - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} \cdot \frac{\cos 4	heta}{1 - \cos 4	heta} = \frac{2\sin^2 4	heta}{\cos 8	heta} \cdot \frac{\cos 4	heta}{2\sin^2 2	heta}$.Using $\sin 4	heta = 2\sin 2	heta \cos 2	heta$,we get $E = \frac{2(4\sin^2 2	heta \cos^2 2	heta)}{\cos 8	heta} \cdot \frac{\cos 4	heta}{2\sin^2 2	heta} = \frac{4\cos^2 2	heta \cos 4	heta}{\cos 8	heta}$.Let $t = 	an^2 2	heta$. Then $\cos 4	heta = \frac{1-t}{1+t}$ and $\cos 8	heta = \frac{1 - 	an^2 4	heta}{1 + 	an^2 4	heta} = \frac{1 - (\frac{2\sqrt{t}}{1-t})^2}{1 + (\frac{2\sqrt{t}}{1-t})^2} = \frac{(1-t)^2 - 4t}{(1-t)^2 + 4t} = \frac{1 - 6t + t^2}{1 + 2t + t^2}$.Also $\cos^2 2	heta = \frac{1}{1+t}$.Substituting these: $E = \frac{4}{1+t} \cdot \frac{1-t}{1+t} \cdot \frac{1+2t+t^2}{1-6t+t^2} = \frac{4(1-t)(1+t)^2}{(1+t)^2(1-6t+t^2)} = \frac{4-4t}{1-6t+t^2}$.Comparing with $\frac{a+bt}{1+ct+dt^2}$,we get $a=4, b=-4, c=-6, d=1$.Thus,$a-b+c-d = 4 - (-4) + (-6) - 1 = 4 + 4 - 6 - 1 = 1$.

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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