Answer
$\dfrac{1}{10}(4\sqrt 3-3)$
Work Step by Step
Let us consider $\cos \theta=\dfrac{3}{5}$
Since, $\sin \theta$ is negative in second quadrant. so, we have $\sin \theta=\sqrt{1-\cos^2 \theta}=-\dfrac{4}{5}$
and $\tan \phi=-\sqrt 3$
This gives: $\phi=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$
Now, $\cos(\theta-\phi)=\cos \theta \cos \phi-\sin \theta \sin \phi$
This implies that
$\cos \theta \cos \phi-\sin \theta \sin \phi=(\dfrac{3}{5})(\dfrac{-1}{2})-(\dfrac{-4}{5})(\dfrac{\sqrt 3}{2})$
or, $=\dfrac{1}{10}(4\sqrt 3-3)$
