If $\tan \alpha =2\sqrt{2},$ then the value of $\frac{\tan \alpha }{\frac{{\sin }^3\alpha }{\cos \alpha }+\sin \alpha .\cos \alpha }$ is

Simplifying the given expression we get

$\frac{\frac{\sin \alpha }{\cos \alpha }}{{\sin }^2\alpha \tan \alpha +\sin \alpha \cos \alpha }$

$=\frac{\sin \alpha }{(\sin \alpha \cos \alpha )(\sin \alpha \tan \alpha +\cos \alpha )}$ ...(taking $\sin \alpha$ common from the denominator)

$=\frac{1}{\sin \alpha \tan \alpha \cos \alpha +{\cos }^2\alpha }$

$=\frac{1}{{\sin }^2\alpha +{\cos }^2\alpha }$

$=1$

Hence answer is D