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tan-2-20-tan-2-40-tan-2-80-33-

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Question Number 41691 by avishek last updated on 11/Aug/18
tan^2 20+tan^2 40+tan^2 80=33
$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33} \\ $$
Answered by ajfour last updated on 12/Aug/18
let tan 20°=m tan 3θ=((3tan θ−tan^3 θ)/(1−3tan^2 θ)) for θ=20° we therefore have (√3)=((3m−m^3 )/(1−3m^2 ))=((m((√3)−m)((√3)+m))/((1−m(√3))(1+m(√3)))) -(i) or m^3 −3(√3)m^2 +(√3) = 3m ..(ii) Also (√3)(1−3m^2 )=m(3−m^2 ) ⇒ m^2 =(((√3)−3m)/(3(√3)−m)) ...(iii) l.h.s.= tan^2 20°+tan^2 (60°−20°)°+tan^2 (60°+20°) =m^2 +((((√3)−m)/(1+m(√3))))^2 +((((√3)+m)/(1−m(√3))))^2 =m^2 +((((√3)−m)/(1+m(√3)))−(((√3)+m)/(1−m(√3))))^2 +((2(√3))/m) [see (i)] =m^2 +(((8m)/(1−3m^2 )))^2 +((2(√3))/m) = m^2 +[((8m)/(1−3((((√3)−3m)/(3(√3)−m)))))]^2 +((2(√3))/m) [see (iii)] =m^2 +(3(√3)−m)^2 +((2(√3))/m) = ((2(m^3 −3(√3)m^2 +(√3))+27m)/m) = ((6m+27m)/m) = 33 [ see (ii)] .
$${let}\:\mathrm{tan}\:\mathrm{20}°={m} \\ $$$$\mathrm{tan}\:\mathrm{3}\theta=\frac{\mathrm{3tan}\:\theta−\mathrm{tan}\:^{\mathrm{3}} \theta}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \theta} \\ $$$${for}\:\:\theta=\mathrm{20}°\:\:{we}\:{therefore}\:{have} \\ $$$$\sqrt{\mathrm{3}}=\frac{\mathrm{3}{m}−{m}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{m}^{\mathrm{2}} }=\frac{{m}\left(\sqrt{\mathrm{3}}−{m}\right)\left(\sqrt{\mathrm{3}}+{m}\right)}{\left(\mathrm{1}−{m}\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+{m}\sqrt{\mathrm{3}}\right)}\:-\left({i}\right) \\ $$$${or}\:\:\:\:\:{m}^{\mathrm{3}} −\mathrm{3}\sqrt{\mathrm{3}}{m}^{\mathrm{2}} +\sqrt{\mathrm{3}}\:=\:\mathrm{3}{m}\:\:\:..\left({ii}\right) \\ $$$${Also}\:\:\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{3}{m}^{\mathrm{2}} \right)={m}\left(\mathrm{3}−{m}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\boldsymbol{{m}}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{3}\boldsymbol{{m}}}{\mathrm{3}\sqrt{\mathrm{3}}−\boldsymbol{{m}}}\:\:\:\:\:…\left({iii}\right) \\ $$$$\:{l}.{h}.{s}.=\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{20}°+\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{60}°−\mathrm{20}°\right)°+\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{60}°+\mathrm{20}°\right) \\ $$$$\:\:={m}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}−{m}}{\mathrm{1}+{m}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}+{m}}{\mathrm{1}−{m}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \\ $$$$\:\:={m}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}−{m}}{\mathrm{1}+{m}\sqrt{\mathrm{3}}}−\frac{\sqrt{\mathrm{3}}+{m}}{\mathrm{1}−{m}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{m}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{see}\:\left({i}\right)\right] \\ $$$$\:={m}^{\mathrm{2}} +\left(\frac{\mathrm{8}{m}}{\mathrm{1}−\mathrm{3}{m}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{m}} \\ $$$$=\:{m}^{\mathrm{2}} +\left[\frac{\mathrm{8}{m}}{\mathrm{1}−\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{3}{m}}{\mathrm{3}\sqrt{\mathrm{3}}−{m}}\right)}\right]^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{m}} \\ $$$$\:\:\:\:\:\left[{see}\:\left({iii}\right)\right] \\ $$$$={m}^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{3}}−{m}\right)^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{m}} \\ $$$$=\:\frac{\mathrm{2}\left({m}^{\mathrm{3}} −\mathrm{3}\sqrt{\mathrm{3}}{m}^{\mathrm{2}} +\sqrt{\mathrm{3}}\right)+\mathrm{27}{m}}{{m}} \\ $$$$=\:\frac{\mathrm{6}{m}+\mathrm{27}{m}}{{m}}\:=\:\mathrm{33}\:\:\:\:\:\left[\:{see}\:\left({ii}\right)\right]\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Aug/18
9∝=Π tan9∝=((9c_1 t−9c_3 t^3 +9c_5 t^5 −9c_7 t^7 −9c_9 t^9 )/(1−9c_2 t^2 +9c_4 t^4 −9c_6 t^6 +9c_8 t^8 )) 9t−((9×8×7)/(3×2))t^3 +((9×8×7×6)/(4×3×2))t^5 −((9×8)/2)t^7 +t^9 =0 9t−4t^3 +126t^5 −36t^7 +t^9 =0 9−4t^2 +126t^4 −36t^6 +t^8 =0 t^8 −36t^6 +126t^4 −4t^2 +9=0 t=tanα x=t^2 x^4 −36x^3 +126x^2 −4x+9=0 tan^2 20+tan^2 40+tan^2 60+tan^2 80=36 tan^2 20+tan^2 40+tan^2 80=36−3=33
$$\mathrm{9}\propto=\Pi \\ $$$${tan}\mathrm{9}\propto=\frac{\mathrm{9}{c}_{\mathrm{1}} {t}−\mathrm{9}{c}_{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{9}{c}_{\mathrm{5}} {t}^{\mathrm{5}} −\mathrm{9}{c}_{\mathrm{7}} {t}^{\mathrm{7}} −\mathrm{9}{c}_{\mathrm{9}} {t}^{\mathrm{9}} }{\mathrm{1}−\mathrm{9}{c}_{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{9}{c}_{\mathrm{4}} {t}^{\mathrm{4}} −\mathrm{9}{c}_{\mathrm{6}} {t}^{\mathrm{6}} +\mathrm{9}{c}_{\mathrm{8}} {t}^{\mathrm{8}} } \\ $$$$\mathrm{9}{t}−\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{3}×\mathrm{2}}{t}^{\mathrm{3}} +\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}}{\mathrm{4}×\mathrm{3}×\mathrm{2}}{t}^{\mathrm{5}} −\frac{\mathrm{9}×\mathrm{8}}{\mathrm{2}}{t}^{\mathrm{7}} +{t}^{\mathrm{9}} =\mathrm{0} \\ $$$$\mathrm{9}{t}−\mathrm{4}{t}^{\mathrm{3}} +\mathrm{126}{t}^{\mathrm{5}} −\mathrm{36}{t}^{\mathrm{7}} +{t}^{\mathrm{9}} =\mathrm{0} \\ $$$$\mathrm{9}−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{126}{t}^{\mathrm{4}} −\mathrm{36}{t}^{\mathrm{6}} +{t}^{\mathrm{8}} =\mathrm{0} \\ $$$${t}^{\mathrm{8}} −\mathrm{36}{t}^{\mathrm{6}} +\mathrm{126}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$${t}={tan}\alpha \\ $$$${x}={t}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{36}{x}^{\mathrm{3}} +\mathrm{126}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{9}=\mathrm{0} \\ $$$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{60}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{36} \\ $$$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{36}−\mathrm{3}=\mathrm{33} \\ $$$$ \\ $$
Commented by ajfour last updated on 13/Aug/18
great, Tanmay Sir.
$${great},\:{Tanmay}\:{Sir}. \\ $$

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