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Question Number 131302 by EDWIN88 last updated on 03/Feb/21
Given sin ((π/6)+tan^(−1) (x))= ((13)/(14)) If tan^(−1) (x)= tan^(−1) (((a(√3))/b)) then ((a+b)/2) =?
$${Given}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)=\:\frac{\mathrm{13}}{\mathrm{14}} \\ $$$$\:{If}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}\sqrt{\mathrm{3}}}{{b}}\right)\:{then}\: \\ $$$$\:\frac{{a}+{b}}{\mathrm{2}}\:=?\: \\ $$
Answered by mr W last updated on 03/Feb/21
t=tan^(−1) x <(π/3) ! sin ((π/6)+t)=((13)/(14)) (1/2)cos t+((√3)/2)sin t=((13)/(14)) cos t+(√3)sin t=((13)/7) u=cos t (√(3(1−u^2 )))=((13)/7)−u 3(1−u^2 )=((169)/(49))+u^2 −((26)/7)u 2u^2 −((13)/7)u+((11)/(49))=0 u=(1/4)(((13±9)/7))=((11)/(14)), (1/7) tan t=(√((1/(cos^2 t))−1))=((5(√3))/(11)), 4(√3)>(√3) ! rejected ⇒t=tan^(−1) ((5(√3))/(11)) tan^(−1) x=t=tan^(−1) ((5(√3))/(11)) ⇒a=5, b=11 ⇒((a+b)/2)=8
$${t}=\mathrm{tan}^{−\mathrm{1}} {x}\:<\frac{\pi}{\mathrm{3}}\:! \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+{t}\right)=\frac{\mathrm{13}}{\mathrm{14}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:{t}=\frac{\mathrm{13}}{\mathrm{14}} \\ $$$$\mathrm{cos}\:{t}+\sqrt{\mathrm{3}}\mathrm{sin}\:{t}=\frac{\mathrm{13}}{\mathrm{7}} \\ $$$${u}=\mathrm{cos}\:{t} \\ $$$$\sqrt{\mathrm{3}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}=\frac{\mathrm{13}}{\mathrm{7}}−{u} \\ $$$$\mathrm{3}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\frac{\mathrm{169}}{\mathrm{49}}+{u}^{\mathrm{2}} −\frac{\mathrm{26}}{\mathrm{7}}{u} \\ $$$$\mathrm{2}{u}^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{7}}{u}+\frac{\mathrm{11}}{\mathrm{49}}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{13}\pm\mathrm{9}}{\mathrm{7}}\right)=\frac{\mathrm{11}}{\mathrm{14}},\:\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\mathrm{tan}\:{t}=\sqrt{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{t}}−\mathrm{1}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{11}},\:\mathrm{4}\sqrt{\mathrm{3}}>\sqrt{\mathrm{3}}\:!\:{rejected} \\ $$$$\Rightarrow{t}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}={t}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\Rightarrow{a}=\mathrm{5},\:{b}=\mathrm{11} \\ $$$$\Rightarrow\frac{{a}+{b}}{\mathrm{2}}=\mathrm{8} \\ $$

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