If $0\le \theta \le 2\mathrm{\pi }$, solve the equation:

$4\tan \left(\theta \right)+2=2\tan \left(\theta \right)$

Solve the following.

$\frac{2 \sin {22}^{\circ }}{\cos {68}^{\circ }}-\frac{2 \cot {22}^{\circ }}{5 \tan {15}^{\circ }}-\frac{8 \tan {45}^{\circ }\tan {20}^{\circ }\tan {40}^{\circ }\tan {50}^{\circ }\tan {70}^{\circ }}{5}$

If $2{\cos }^2\mathrm{\theta }-5\mathrm{cos\theta }+2=0$,  $0°<\mathrm{\theta }<90°$, then the value of $(\mathrm{sec\theta }+\mathrm{tan\theta })$ is

If $\cos \left(2\mathrm{\theta }+54°\right)=\mathrm{sin\theta }$, $0°<\left(2\mathrm{\theta }+54°\right)<90°$, then what is the value of $\frac{1}{\cot 5\theta +\sec \frac{5\theta }{2}}?$