Given-that-tan-and-tan-are-the-roots-of-the-equation-x-2-3ax-4a-1-0-where-a-gt-1-and-pi-2-pi-2-Evaluate-tan-2-

Question Number 156426 by ZiYangLee last updated on 11/Oct/21

Given that \tan α and \tan β are the roots

of the equation x^{2} + 3 a x + 4 a + 1 = 0,

where a > 1 and α, β \in (- \frac{π}{2}, \frac{π}{2}) .

Evaluate \tan (\frac{α + β}{2}) .

Answered by mr W last updated on 11/Oct/21

\tan α + \tan β = - 3 a

\tan α \tan β = 4 a + 1

\tan (α + β) = \frac{- 3 a + 4 a + 1}{1 + 3 a (4 a + 1)} = \frac{a + 1}{12 a^{2} + 3 a + 1}

= \frac{2 \tan \frac{α + β}{2}}{1 - \tan^{2} \frac{α + β}{2}} = \frac{2 t}{1 - t^{2}}

\frac{2 t}{1 - t^{2}} = \frac{a + 1}{12 a^{2} + 3 a + 1}

(a + 1) t^{2} + 2 (12 a^{2} + 3 a + 1) t - (a + 1) = 0

t = \tan \frac{α + β}{2} = - \frac{(12 a^{2} + 3 a + 1) \pm \sqrt{{(12 a^{2} + 3 a + 1)}^{2} + {(a + 1)}^{2}}}{a + 1}

Facebook Tweet Pin