Let f:(-1,1) rightarrow mathbb{R} be such tha | vedclass

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(AB) Given $f(\cos 4 	heta) = \frac{2}{2-\sec^2 	heta}$.We know that $\sec^2 	heta = \frac{1}{\cos^2 	heta} = \frac{2}{1+\cos 2 	heta}$.Substitut

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$1+\sqrt{\frac{2}{3}}$

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(AB) Given $f(\cos 4 	heta) = \frac{2}{2-\sec^2 	heta}$.We know that $\sec^2 	heta = \frac{1}{\cos^2 	heta} = \frac{2}{1+\cos 2 	heta}$.Substituting this into the expression for $f(\cos 4 	heta)$:$f(\cos 4 	heta) = \frac{2}{2 - \frac{2}{1+\cos 2 	heta}} = \frac{2(1+\cos 2 	heta)}{2(1+\cos 2 	heta) - 2} = \frac{1+\cos 2 	heta}{\cos 2 	heta} = 1 + \frac{1}{\cos 2 	heta}$.Now,let $\cos 4 	heta = \frac{1}{3}$.Using the identity $\cos 4 	heta = 2 \cos^2 2 	heta - 1$,we have $2 \cos^2 2 	heta - 1 = \frac{1}{3} \Rightarrow 2 \cos^2 2 	heta = \frac{4}{3} \Rightarrow \cos^2 2 	heta = \frac{2}{3}$.Thus,$\cos 2 	heta = \pm \sqrt{\frac{2}{3}}$.Substituting this back into the expression for $f(\cos 4 	heta)$:$f\left(\frac{1}{3}ight) = 1 + \frac{1}{\pm \sqrt{2/3}} = 1 \pm \sqrt{\frac{3}{2}}$.Therefore,the values are $1+\sqrt{\frac{3}{2}}$ and $1-\sqrt{\frac{3}{2}}$.

Let $f:(-1,1) \rightarrow \mathbb{R}$ be such that $f(\cos 4 \theta) = \frac{2}{2-\sec^2 \theta}$ for $\theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value$(s)$ of $f\left(\frac{1}{3}\right)$ is (are):

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