If tan-1- tan-2- tan-3 are the real roots of x 3- a -1 x 2- b - a x - b -0 where -1- -2- -3 are acute then -1-2-3-A- 5 -6B- -2C- 6 -5D- -4

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Quadratic Equation

If tanθ1, tan...

Question

If $t a n θ_{1}, t a n θ_{2}, t a n θ_{3}$ are the real roots of $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0 w h e r e θ_{1}, θ_{2}, θ_{3}$ are acute then $θ_{1} + θ_{2} + θ_{3} =$

$\frac{π}{2}$

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$\frac{5 π}{6}$

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$\frac{6 π}{5}$

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$\frac{π}{4}$

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Solution

The correct option is D
$\frac{π}{4}$

$t a n θ_{1} + t a n θ_{2} + t a n θ_{3} = a + 1, t a n θ_{1} t a n θ_{2} t a n θ_{3} = b t a n θ_{1} t a n θ_{2} + t a n θ_{2} t a n θ_{3} + t a n θ_{3} t a n θ_{1} = b - a$
$N o w t a n (θ_{1} + θ_{2} + θ_{3}) = \frac{t a n θ_{1} + t a n θ_{2} + t a n θ_{3} - t a n θ_{1} t a n θ_{2} t a n θ_{3}}{1 - t a n θ_{1} t a n θ_{2} - t a n θ_{2} t a n θ_{3} - t a n θ_{3} t a n θ_{1}} = \frac{a + 1 - b}{1 - (b - a)} = \frac{a - b + 1}{1 - b + a} = 1 ∴ θ_{1} + θ_{2} + θ_{3} = \frac{π}{4}$

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If tan θ1, tan θ2, tan θ3 are the real roots of x3−(a+1)x2+(b−a)x−b=0 where θ1, θ2, θ3 are acute then θ1+θ2+θ3=

If $t a n θ_{1}, t a n θ_{2}, t a n θ_{3}$ are the real roots of $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0 w h e r e θ_{1}, θ_{2}, θ_{3}$ are acute then $θ_{1} + θ_{2} + θ_{3} =$