If $\int \frac{1}{5+4\cos 2\theta }d\theta =A{\tan }^{-1}\left(B\left(\tan \theta \right)\right)+c$ then $(A,B)=$

$\int \frac{1}{5+4\cos 2\theta }d\theta$

$Put\tan \theta =t\Rightarrow {\sec }^2\theta d\theta =dt$

$(1+{\tan }^2\theta )d\theta =dt\Rightarrow d\theta =\frac{dt}{1+t^2}$

$=\int \frac{1}{\left[5+4\left(\frac{1-t^2}{1+t^2}\right)\right]}\frac{dt}{(1+t^2)}$

$=\int \frac{dt}{5+5t^2+4-4t^2}$

$=\int \frac{dt}{t^2+9}$

$=\frac{1}{3}{\tan }^{-1}\left(\frac{{\displaystyle t}}{{\displaystyle 3}}\right)+c$

$=\frac{1}{3}{\tan }^{-1}\left(\frac{{\displaystyle \tan \theta }}{{\displaystyle 3}}\right)+c$

$\therefore A=\frac{1}{3},B=\frac{1}{3}$