If  $\alpha$ is the negative real root of the equation $x^3+b{x}^2+cx+1=0(b<c)$ then the value of  ${\tan }^{-1}\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)=$

$x^3+b{x}^2+cx+1=0$
$f\left(x\right)=x^3+b{x}^2+cx+1$
$\text{ Now, }f\left(0\right)=1>0,f(-1)=b-c<0$
$\text{ so, the function }f\left(x\right)\text{ has a roots in }-1,\alpha <0$
$\text{ Now, }{\tan }^{-1}\left(\alpha \right)+{\tan }^{-1}(1/\alpha )$
$\begin{array}{l}={\tan }^{-1}\alpha -\pi +{\cot }^{-1}\alpha \\ =-\pi +\left({\tan }^{-1}\alpha +{\cot }^{-1}\alpha \right)\\ =-\pi +\pi /2=-\pi /2\end{array}$