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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 +3ax+4a+1=0, where a>1 and α,β∈(−(π/2),(π/2)). Evaluate tan(((α+β)/2)).
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{tan}\:\alpha\:\mathrm{and}\:\mathrm{tan}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} +\mathrm{3}{ax}+\mathrm{4}{a}+\mathrm{1}=\mathrm{0},\: \\ $$$$\mathrm{where}\:{a}>\mathrm{1}\:\mathrm{and}\:\alpha,\beta\in\left(−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right).\: \\ $$$$\mathrm{Evaluate}\:\mathrm{tan}\left(\frac{\alpha+\beta}{\mathrm{2}}\right). \\ $$
Answered by mr W last updated on 11/Oct/21
tan α+tan β=−3a tan αtan β=4a+1 tan (α+β)=((−3a+4a+1)/(1+3a(4a+1)))=((a+1)/(12a^2 +3a+1)) =((2tan ((α+β)/2))/(1−tan^2 ((α+β)/2)))=((2t)/(1−t^2 )) ((2t)/(1−t^2 ))=((a+1)/(12a^2 +3a+1)) (a+1)t^2 +2(12a^2 +3a+1)t−(a+1)=0 t=tan ((α+β)/2)=−(((12a^2 +3a+1)±(√((12a^2 +3a+1)^2 +(a+1)^2 )))/(a+1))
$$\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta=−\mathrm{3}{a} \\ $$$$\mathrm{tan}\:\alpha\mathrm{tan}\:\beta=\mathrm{4}{a}+\mathrm{1} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{−\mathrm{3}{a}+\mathrm{4}{a}+\mathrm{1}}{\mathrm{1}+\mathrm{3}{a}\left(\mathrm{4}{a}+\mathrm{1}\right)}=\frac{{a}+\mathrm{1}}{\mathrm{12}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}} \\ $$$$=\frac{\mathrm{2tan}\:\frac{\alpha+\beta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha+\beta}{\mathrm{2}}}=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{{a}+\mathrm{1}}{\mathrm{12}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}} \\ $$$$\left({a}+\mathrm{1}\right){t}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{12}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}\right){t}−\left(\mathrm{a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\frac{\alpha+\beta}{\mathrm{2}}=−\frac{\left(\mathrm{12}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}\right)\pm\sqrt{\left(\mathrm{12}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}\right)^{\mathrm{2}} +\left({a}+\mathrm{1}\right)^{\mathrm{2}} }}{{a}+\mathrm{1}} \\ $$

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