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

$-\frac{\pi}{2}$

Let  $f(x)=x^3 + bx^2 + cx + 1$.Then,

 $f(0) =  1 > 0 $ and $ f(-1) = b - c< 0 $       [∵ b < c ]

$⇒ \alpha $ lies between -1 and 0.

$⇒ \alpha < 0 $

$⇒ tan^{-1}\left(\frac{1}{\alpha }\right) = - \pi + cot^{-1} \alpha $

$⇒ tan^{-1} \alpha + tan^{-1}\frac{1}{\alpha }= -\pi + tan^{-1}\alpha + cot^{-1} \alpha $

$⇒ tan^{-1} \alpha + tan^{-1}\frac{1}{\alpha } = - \pi + \frac{\pi}{2} = -\frac{\pi}{2}$