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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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