Let $f: (-1, 1) \to \mathbb{R}$ 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)

$f(\cos 4\theta)=\dfrac{2}{2-\sec^{2}\theta}$ $\theta\in\left(0,\dfrac{\pi}{4}\right)\cup\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$ $f:(-1,1)\to\mathbb{R}$ $\text{To find: } f\left(\dfrac{1}{3}\right)$ $(1)\ 1-\sqrt{\dfrac{3}{2}} \qquad (2)\ 1+\sqrt{\dfrac{3}{2}}$ $(3)\ 1-\sqrt{\dfrac{2}{3}} \qquad (4)\ 1+\sqrt{\dfrac{2}{3}}$ $\text{Plan: simplify RHS using } \cos 2\theta$ $\text{then put } \cos 4\theta=\dfrac{1}{3} \Rightarrow \cos 2\theta = ?$ $\cos 2\theta=2\cos^{2}\theta-1 \Rightarrow \cos^{2}\theta=\dfrac{1+\cos 2\theta}{2}$ $\sec^{2}\theta=\dfrac{1}{\cos^{2}\theta}=\dfrac{2}{1+\cos 2\theta}$ $\dfrac{2}{2-\sec^{2}\theta}=\dfrac{2(1+\cos 2\theta)}{2(1+\cos 2\theta)-2}$ $=\dfrac{1+\cos 2\theta}{\cos 2\theta}=1+\sec 2\theta$ $f(\cos 4\theta)=1+\sec 2\theta$ $\cos 4\theta=2\cos^{2}2\theta-1$ $2\cos^{2}2\theta-1=\dfrac{1}{3} \Rightarrow \cos^{2}2\theta=\dfrac{2}{3}$ $\cos 2\theta=\pm\sqrt{\dfrac{2}{3}},\qquad \sec 2\theta=\pm\sqrt{\dfrac{3}{2}}$ $\theta\in\left(0,\dfrac{\pi}{4}\right):\ 2\theta\in\left(0,\dfrac{\pi}{2}\right)\Rightarrow \cos 2\theta>0$ $f\left(\dfrac{1}{3}\right)=1+\sqrt{\dfrac{3}{2}}$ $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right):\ 2\theta\in\left(\dfrac{\pi}{2},\pi\right)\Rightarrow \cos 2\theta<0$ $f\left(\dfrac{1}{3}\right)=1-\sqrt{\dfrac{3}{2}}$ $\text{Both attained: } 1+\sqrt{\dfrac{3}{2}} \ \text{ and } \ 1-\sqrt{\dfrac{3}{2}}$ $\boxed{f\left(\dfrac{1}{3}\right)=1\pm\sqrt{\dfrac{3}{2}}}\quad\Rightarrow\ (1)\ \text{and}\ (2)$