If x_1, x_2, x_3 are the roots of the equatio | vedclass

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(A) Given the cubic equation $x^3 - x^2 	an 	heta + x 	an^2 	heta + 	an 	heta = 0$.By Vieta's formulas, the sum of roots, sum of roots taken two

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Given the cubic equation $x^3 - x^2 	an 	heta + x 	an^2 	heta + 	an 	heta = 0$.By Vieta's formulas, the sum of roots, sum of roots taken two at a time, and product of roots are:$x_1 + x_2 + x_3 = 	an 	heta$$x_1 x_2 + x_2 x_3 + x_3 x_1 = 	an^2 	heta$$x_1 x_2 x_3 = -	an 	heta$We use the identity $	an^{-1} x_1 + 	an^{-1} x_2 + 	an^{-1} x_3 = 	an^{-1} \left( \frac{(x_1 + x_2 + x_3) - x_1 x_2 x_3}{1 - (x_1 x_2 + x_2 x_3 + x_3 x_1)} ight)$.Substituting the values:$= 	an^{-1} \left( \frac{	an 	heta - (-	an 	heta)}{1 - 	an^2 	heta} ight) = 	an^{-1} \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight)$.Using the double angle formula $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$, we get:$= 	an^{-1} (	an 2	heta) = 2	heta$.At $	heta = \frac{\pi}{12}$, the value is $2 	imes \frac{\pi}{12} = \frac{\pi}{6}$.Thus, option $A$ is correct.

If $x_1, x_2, x_3$ are the roots of the equation $x^3 - x^2 \tan \theta + x \tan^2 \theta + \tan \theta = 0$ and $0 < \theta < \frac{\pi}{4}$, then the value of $\tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3$ at $\theta = \frac{\pi}{12}$ is

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