Given that tan α and tan β are the roots of the equation x^2 +3ax+4a+1=0, where a>1 and α,β∈(−(π/2),(π/2)). Evaluate tan(((α+β)/2)).


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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

	$\tan α + \tan β = - 3 a$ $\tan α \tan β = 4 a + 1$ $\tan (α + β) = \frac{- 3 a + 4 a + 1}{1 + 3 a (4 a + 1)} = \frac{a + 1}{12 a^{2} + 3 a + 1}$ $= \frac{2 \tan \frac{α + β}{2}}{1 - \tan^{2} \frac{α + β}{2}} = \frac{2 t}{1 - t^{2}}$ $\frac{2 t}{1 - t^{2}} = \frac{a + 1}{12 a^{2} + 3 a + 1}$ $(a + 1) t^{2} + 2 (12 a^{2} + 3 a + 1) t - (a + 1) = 0$ $t = \tan \frac{α + β}{2} = - \frac{(12 a^{2} + 3 a + 1) \pm \sqrt{{(12 a^{2} + 3 a + 1)}^{2} + {(a + 1)}^{2}}}{a + 1}$
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