If $\cos \theta =\frac{3}{5}$ and $\theta$ is not in the first quadrant, then $\frac{5\tan \left(\pi +\theta \right)+4\cos \left(\pi +\theta \right)}{5\sec \left(2\pi -\theta \right)-4\cot \left(2\pi +\theta \right)}=$

$\theta$ lies in fourth quadrant substitute corresponding values

$$\begin{gathered} \frac{5\tan (\pi + \theta )+ 4 \cos (\pi + \theta )}{5 sec ( 2 \pi - \theta )-4 cot( 2 \pi + \theta )}=\frac{5\tan \theta -4\cos \theta }{5sec\theta -4cot\theta } \\ =\frac{5\left({\displaystyle \frac{4}{3}}\right)-4\left({\displaystyle \frac{3}{5}}\right)}{5\left({\displaystyle \frac{5}{3}}\right)-4\left({\displaystyle \frac{3}{4}}\right)} \\ =\frac{4}{5} \end{gathered}$$