If $0<\theta <\frac{\pi }{4}$ , and $f\left(\theta \right)=\left(1+\cot \theta -\cos ec\theta \right)\left(1+\tan \theta +\sec \theta \right)$ , then $f\left(\frac{\pi }{6}\right)=$  

$=\left(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }\right)\left(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\right)$

$=\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)\left(\frac{\sin \theta +\cos \theta +1}{\cos \theta }\right)$

$=\frac{{\left(\sin \theta +\cos \theta \right)}^2-1}{\sin \theta \cos \theta }$

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

$=2$

$\therefore$ $f\left(\frac{\pi }{6}\right)=2$