The simplest form of frac{2}{sqrt{2+sqrt{2+sq | vedclass

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(A) Given expression: $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4x}}}}$Using the identity $1 + \cos 2	heta = 2 \cos^2 	heta$,we have $2 + 2 \cos 4x =

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$\sec \frac{x}{2}$

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(A) Given expression: $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4x}}}}$Using the identity $1 + \cos 2	heta = 2 \cos^2 	heta$,we have $2 + 2 \cos 4x = 2(1 + \cos 4x) = 2(2 \cos^2 2x) = 4 \cos^2 2x$.Substituting this into the innermost radical: $\sqrt{2+2 \cos 4x} = \sqrt{4 \cos^2 2x} = 2 \cos 2x$.Now the expression becomes: $\frac{2}{\sqrt{2+\sqrt{2+2 \cos 2x}}}$.Again,using $2 + 2 \cos 2x = 2(1 + \cos 2x) = 2(2 \cos^2 x) = 4 \cos^2 x$.Substituting this: $\sqrt{2+2 \cos 2x} = \sqrt{4 \cos^2 x} = 2 \cos x$.Now the expression becomes: $\frac{2}{\sqrt{2+2 \cos x}}$.Using $2 + 2 \cos x = 2(1 + \cos x) = 2(2 \cos^2 \frac{x}{2}) = 4 \cos^2 \frac{x}{2}$.Thus,$\sqrt{2+2 \cos x} = 2 \cos \frac{x}{2}$.Finally,the expression simplifies to $\frac{2}{2 \cos \frac{x}{2}} = \frac{1}{\cos \frac{x}{2}} = \sec \frac{x}{2}$.

The simplest form of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4x}}}}$ is

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