Question

If
$x_1,x_2,x_3$ are the roots of $x^3-6x^2+11x-6=0$ ,then
$co{t}^{-1}\left(x_1\right)+co{t}^{-1}\left(x_2\right)+co{t}^{-1}\left(x_3\right)=$

• A. 0
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{2}$

$x^3-6x^2+11x-6=0$
$(x-1)(x^2-5x+6)=0$
$(x-1)(x-2)(x-3)=0$
$x_1=1$ ,$x_2=2$ and $x_3=3$
Hence
$co{t}^{-1}\left(x_1\right)+co{t}^{-1}\left(x_2\right)+co{t}^{-1}\left(x_3\right)$
$=co{t}^{-1}\left(1\right)+co{t}^{-1}\left(2\right)+co{t}^{-1}\left(3\right)$
$=\frac{\pi }{4}+ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{3}\right)$
$=\frac{\pi }{4}+ta{n}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)$
$=\frac{\pi }{4}+ta{n}^{-1}\left(\frac{5}{6-1}\right)$
$=\frac{\pi }{4}+ta{n}^{-1}\left(1\right)$
$=\frac{\pi }{4}+\frac{\pi }{4}$
$=\frac{\pi }{2}$