If the equation `x^3+b x^2+c x+1=0,(bltc),` has only one real root `alpha` , then the value of `2tan^(-1)(cosec alpha)+tan^(-1)(2 sinalpha sec^2alpha)` is

Options

A. `-pi`
B. `-(pi)/(2)`
C. `(pi)/(2)`
D. `pi`

Answer

A

Explanation

Let `f(x) = x^(3) + bx^(2) + cx + 1`
`f(0) = 1 gt 0, f(-1) = b - c lt 0`
So, `alpha in (-1, 0)`
So, `2 tan^(-1) (cosec alpha) + tan^(-1) (2 sin alpha sec^(2) alpha)`
`= 2 tan^(-1) ((1)/(sin alpha)) + tan^(-1) ((2 sinalpha)/(1 - sin^(2) alpha))`
`= 2 [tan^(-1) ((1)/(sin alpha)) + tan^(-1) (sin alpha)]`
`= 2 (-(pi)/(2)) = -pi` (as `sin alpha lt 0`)