The value of tan 3a cot a cannot lie in

$\begin{array}{r}\tan 3\alpha \cot \alpha =\frac{3\tan \alpha -{\tan }^3\alpha }{\tan \alpha \left(1-3{\tan }^2\alpha \right)}\\ =\frac{3-{\tan }^2\alpha }{1-3{\tan }^2\alpha }=x\text{ (say) }\\ {\tan }^2\alpha =\frac{x-3}{3x-1}=\frac{(3x-1)(x-3)}{(3x-1{)}^2}\end{array}$

 Since  ${\tan }^2\alpha$ is non-negative, either    $x<1/3\text{ or }x\ge 3,$  so x cannot lie between 1/3 and 3.