If 3 $\cos$ $\theta$ = 1, find the value of $\frac{6{\sin }^2\theta +{\tan }^2\theta }{4\cos \theta }$

Given,

$\cos \theta =\frac{1}{3}$

$\Rightarrow \sin \theta =\frac{2\sqrt{2}}{3}$

$\Rightarrow \tan \theta =2\sqrt{2}$

$\frac{6{\sin }^2\theta +{\tan }^2\theta }{4\cos \theta }$

$=\frac{6\times \frac{8}{9}+8}{4\times \frac{1}{3}}$

upon solving the above equation, we get,

$=10$