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

$$\begin{gathered} At x=-1, the expression x^3+b{x}^2+cx+1=b-c<0 \left(\because b<c\right) \\ At x=0, the expression x^3+b{x}^2+cx+1=1>0 \\ \Rightarrow the equation x^3+b{x}^2+cx+1==0 must have a real root between -1 and 0 \\ \Rightarrow the only real root \alpha of the equation must lie between -1 and 0 \\ \Rightarrow \alpha <0 \\ \Rightarrow {\tan }^{-1}\left(\frac{1}{\alpha }\right)=-\mathrm{\pi }+{\cot }^{-1}\mathrm{\alpha } \\ \therefore {\tan }^{-1}\mathrm{\alpha }+{\tan }^{-1}\left(\frac{1}{\mathrm{\alpha }}\right)={\tan }^{-1}\mathrm{\alpha }-\mathrm{\pi }+{\cot }^{-1}\mathrm{\alpha } \\ =-\mathrm{\pi }+\frac{\mathrm{\pi }}{2} \\ =-\frac{\mathrm{\pi }}{2} \\ = -1.57 \end{gathered}$$