$\frac{{\sec }^2\theta -\tan \theta }{{\sec }^2\theta +\tan \theta }$$liesbetween$$\left[b,a\right], then 3b+a=$

$\begin{array}{l}Lety=\frac{{\sec }^2\theta -\tan \theta }{{\sec }^2\theta +\tan \theta }\\ y=\frac{1+{\tan }^2\theta -\tan \theta }{1+{\tan }^2\theta +\tan \theta }\\ \Rightarrow y+y{\tan }^2\theta +y\tan \theta =1+{\tan }^2\theta -\tan \theta \\ \Rightarrow (y-1){\tan }^2\theta +(y+1)\tan \theta +(y-1)=0\end{array}$

Since tan$\theta$ is real, we have $b^2-4ac\ge 0$

$\Rightarrow {(y+1)}^2-4{(y-1)}^2\ge 0$

$\Rightarrow 4{(y-1)}^2-{(y+1)}^2\le 0$

$\Rightarrow 3y^2-10y+3\le 0$

$\Rightarrow \frac{1}{3}\le y\le 3$

$$\begin{gathered} Here b=\frac{1}{3},a=3 \\ \therefore 3b+a=4 \end{gathered}$$