If ${\tan }^2\theta =2{\tan }^2\varphi +1,\text{ then }\cos 2\theta +{\sin }^2\varphi$ equals

We have,

$\begin{array}{l}{\tan }^2\theta =2{\tan }^2\varphi +1\\ =1+{\tan }^2\theta =2\left(1+{\tan }^2\varphi \right)\\ ={\sec }^2\theta =2{\sec }^2\varphi \\ ={\cos }^2\varphi =2{\cos }^2\theta \\ ={\cos }^2\varphi =1+\cos 2\theta \\ \Rightarrow \cos 2\theta ={\cos }^2\varphi -1\\ \Rightarrow \cos 2\theta =-{\sin }^2\varphi \Rightarrow {\sin }^2\varphi +\cos 2\theta =0\end{array}$