If $\varphi ={\tan }^{-1}\left(\frac{1}{\sqrt{\cos 2\theta }}\right)-{\tan }^{-1} \left(\sqrt{\cos 2\theta }\right)$ ; then show that ${\tan }^2\varphi =\frac{{\sin }^4\theta }{\cos 2\theta }\mathrm{and} \sec \varphi =\pm \frac{1}{2} \left(\sqrt{\cos 2\theta }\right)+\left(\sqrt{\sec 2\theta }\right)$