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If $x_1, x_2, x_3$ are the real 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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Verified Answer
The correct answer is:
$\frac{\pi}{6}$
According to given informations,
$\begin{aligned} & x_1+x_2+x_3=\tan \theta \\ & x_1 x_2+x_2 x_3+x_3 x_4=\tan ^2 \theta \\ & \text { and } \quad x_1 x_2 x_3=-\tan \theta \\ & \text { So, } \tan ^{-1} x_1+\tan ^{-1} x_2+\tan ^{-1} x_3\end{aligned}$
$\begin{aligned} & =\tan ^{-1}\left(\frac{\left(x_1+x_2+x_3\right)-x_1 x_2 x_3}{1-\left(x_1 x_2+x_2 x_3+x_3 x_4\right)}\right) \\ & =\tan ^{-1}\left(\frac{\tan \theta+\tan \theta}{1-\tan ^2 \theta}\right) \\ & =\tan ^{-1} \tan 2 \theta=2 \theta\end{aligned}$
At $\quad \theta=\frac{\pi}{12}$
$\tan ^{-1} x_1+\tan ^{-1} x_2+\tan ^{-1} x_3=\frac{\pi}{6}$
Hence, option (a) is correct.
$\begin{aligned} & x_1+x_2+x_3=\tan \theta \\ & x_1 x_2+x_2 x_3+x_3 x_4=\tan ^2 \theta \\ & \text { and } \quad x_1 x_2 x_3=-\tan \theta \\ & \text { So, } \tan ^{-1} x_1+\tan ^{-1} x_2+\tan ^{-1} x_3\end{aligned}$
$\begin{aligned} & =\tan ^{-1}\left(\frac{\left(x_1+x_2+x_3\right)-x_1 x_2 x_3}{1-\left(x_1 x_2+x_2 x_3+x_3 x_4\right)}\right) \\ & =\tan ^{-1}\left(\frac{\tan \theta+\tan \theta}{1-\tan ^2 \theta}\right) \\ & =\tan ^{-1} \tan 2 \theta=2 \theta\end{aligned}$
At $\quad \theta=\frac{\pi}{12}$
$\tan ^{-1} x_1+\tan ^{-1} x_2+\tan ^{-1} x_3=\frac{\pi}{6}$
Hence, option (a) is correct.
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