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

$\cos 4\mathrm{\theta }=\frac{1}{3}\Rightarrow \cos 2\mathrm{\theta }=\pm \sqrt{\frac{1+\cos 4\mathrm{\theta }}{2}}=\pm \sqrt{\frac{2}{3}}$

$$\begin{gathered} \mathrm{f}\left(\cos 4\mathrm{\theta }\right)=\frac{2}{2-{\sec }^2\mathrm{\theta }} \mathrm{for} \mathrm{\theta }\in \left(0,\frac{\mathrm{\pi }}{4}\right)\cup \left(\frac{\mathrm{\pi }}{4}, \frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \mathrm{f}\left(\cos 4\mathrm{\theta }\right)=\frac{2{\cos }^2\mathrm{\theta }}{2{\cos }^2\mathrm{\theta }-1}=\frac{1+\cos 2\mathrm{\theta }}{\cos 2\mathrm{\theta }} \\ \Rightarrow \mathrm{f}\left(\cos 4\mathrm{\theta }\right)=1+\frac{1}{\cos 2\mathrm{\theta }}=1\pm \sqrt{\frac{3}{2}} \end{gathered}$$