If the equation x 3 + b x 2 + c x + 1 = 0 ( b c ) has only one real root . Then the value of 2 tan - 1 ( cosec ) + tan - 1 ( 2 sin sec 2 ) is:

Question

If the equation $x^{3} + b x^{2} + c x + 1 = 0 (b < c)$ has only one real root $α$ . Then the value of $2 \tan^{- 1} (cosec α) + \tan^{- 1} (2 \sin α \sec^{2} α)$ is:

Smart Achievers · Accepted Answer

(2)

$x^{3} + b x^{2} + c x + 1 = 0$

$f (- 1) = b - c < 0$

$f (0) = 1 > 0$

$B \in (0, 1)$

$y = - 2 \tan^{- 1} (\cos e c B) - \tan^{- 1} (\frac{2 \sin B}{\cos^{2} B})$

$= - (π + \tan^{- 1} \frac{2 \cos B}{1 - c o s e c^{2} B}) - \tan^{- 1} \frac{2 \sin B}{\cos^{2} B}$

$= - π$

Solution

Q.
If the equation $x^{3} + b x^{2} + c x + 1 = 0 (b < c)$ has only one real root $α$ . Then the value of $2 \tan^{- 1} (cosec α) + \tan^{- 1} (2 \sin α \sec^{2} α)$ is:

1 $- \frac{π}{2}$

2 $- π$

3 $\frac{π}{2}$

4 $π$

Ans.
(2)

$x^{3} + b x^{2} + c x + 1 = 0$

$f (- 1) = b - c < 0$

$f (0) = 1 > 0$

$B \in (0, 1)$

$y = - 2 \tan^{- 1} (\cos e c B) - \tan^{- 1} (\frac{2 \sin B}{\cos^{2} B})$

$= - (π + \tan^{- 1} \frac{2 \cos B}{1 - c o s e c^{2} B}) - \tan^{- 1} \frac{2 \sin B}{\cos^{2} B}$

$= - π$

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Solution

Q. If the equation x3+bx2+cx+1=0 (b<c) has only one real root α. Then the value of 2tan-1(cosec α)+tan-1(2sin α sec2 α) is:

1 -π2

2 -π

3 π2

4 π

Ans. (2) x3+bx2+cx+1=0 f(-1)=b-c<0 f(0)=1>0 B∈(0,1) y=-2tan-1(cosecB)-tan-1(2sinBcos2B) =-(π+tan-12cosB1-cosec2B)-tan-12sinBcos2B =-π