If the equation $x^3+b{x}^2+cx+1=0(b<c)$ has only one root  $\alpha$  then the value of  $2{\tan }^{-1}\left(\cos ec\alpha \right)+{\tan }^{-1}\left(2\sin \alpha {\sec }^2\alpha \right)$ is

$$\begin{gathered} f\left(0\right)=1>0,f(-1)=b-c<0\text{ so, }\alpha \in (-1,0) \\ f\left(0\right)=1>0,f(-1)=b-c<0\text{ so, }\alpha \in (-1,0) \\ 2{\tan }^{-1}\left(\cos ec\alpha \right)+{\tan }^{-1}\left(2\sin \alpha {\sec }^2\alpha \right) \\ =2{\tan }^{-1}\left(\frac{1}{\sin \alpha }\right)+{\tan }^{-1}\left(\frac{2\sin \alpha }{1-{\sin }^2\alpha }\right) \\ =2[{\tan }^{-1}\left(\frac{1}{\sin \alpha }\right)+{\tan }^{-1}\left(\sin \alpha \right)] \\ =2\left(\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=-\pi \end{gathered}$$