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

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Question

(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} =

Vedclass · Accepted Answer

$1$

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(B) Given expression: $\frac{\sec 8	heta - 1}{\sec 4	heta - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} \cdot \frac{\cos 4	heta}{1 - \cos 4	heta} = \frac{2 \sin^2 4	heta}{2 \sin^2 2	heta} \cdot \frac{\cos 4	heta}{\cos 8	heta} = \frac{4 \sin^2 2	heta \cos^2 2	heta}{\sin^2 2	heta} \cdot \frac{\cos 4	heta}{\cos 8	heta} = \frac{4 \cos^2 2	heta \cos 4	heta}{\cos 8	heta}$.Let $t = 	an 2	heta$. We know $\cos 4	heta = \frac{1 - t^2}{1 + t^2}$, $\cos 8	heta = \frac{1 - 	an^2 4	heta}{1 + 	an^2 4	heta} = \frac{1 - \left(\frac{2t}{1-t^2}ight)^2}{1 + \left(\frac{2t}{1-t^2}ight)^2} = \frac{(1-t^2)^2 - 4t^2}{(1-t^2)^2 + 4t^2} = \frac{1 - 6t^2 + t^4}{(1+t^2)^2}$, and $\cos^2 2	heta = \frac{1}{1+t^2}$.Substituting these, we get:$\frac{4 \left(\frac{1}{1+t^2}ight) \left(\frac{1-t^2}{1+t^2}ight)}{\frac{1-6t^2+t^4}{(1+t^2)^2}} = \frac{4(1-t^2)}{1-6t^2+t^4} = \frac{4 - 4 	an^2 2	heta}{1 - 6 	an^2 2	heta + 	an^4 2	heta}$.Comparing with $\frac{a + b 	an^2 2	heta}{1 + c 	an^2 2	heta + d 	an^4 2	heta}$, we get $a=4, b=-4, c=-6, d=1$.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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