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

$$\begin{gathered} \begin{array}{l}\mathrm{Let} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+{\mathrm{bx}}^2+\mathrm{cx}+1=0,(\mathrm{b}<\mathrm{c}) \\ \mathrm{then} \mathrm{f}\left(0\right)=1>0\\ \mathrm{and} \mathrm{f}(-1)=\mathrm{b}-\mathrm{c}<0\\ \therefore \mathrm{f}\left(0\right)\mathrm{f}(-1)<0\end{array} \\ Root \alpha lies between (-1,0) \end{gathered}$$
$\sin ce \alpha <0, {\tan }^{-1}\left(\mathrm{\alpha }\right)+{\tan }^{-1}\left(\frac{1}{\mathrm{\alpha }}\right)=\frac{-\mathrm{\pi }}{2}$