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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

tan220+tan240+tan280=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)] .

lettan20°=mtan3θ=3tanθ−tan3θ1−3tan2θforθ=20°wethereforehave3=3m−m31−3m2=m(3−m)(3+m)(1−m3)(1+m3)−(i)orm3−33m2+3=3m..(ii)Also3(1−3m2)=m(3−m2)⇒m2=3−3m33−m...(iii)l.h.s.=tan220°+tan2(60°−20°)°+tan2(60°+20°)=m2+(3−m1+m3)2+(3+m1−m3)2=m2+(3−m1+m3−3+m1−m3)2+23m[see(i)]=m2+(8m1−3m2)2+23m=m2+[8m1−3(3−3m33−m)]2+23m[see(iii)]=m2+(33−m)2+23m=2(m3−33m2+3)+27mm=6m+27mm=33[see(ii)].

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

9∝=Πtan9∝=9c1t−9c3t3+9c5t5−9c7t7−9c9t91−9c2t2+9c4t4−9c6t6+9c8t89t−9×8×73×2t3+9×8×7×64×3×2t5−9×82t7+t9=09t−4t3+126t5−36t7+t9=09−4t2+126t4−36t6+t8=0t8−36t6+126t4−4t2+9=0t=tanαx=t2x4−36x3+126x2−4x+9=0tan220+tan240+tan260+tan280=36tan220+tan240+tan280=36−3=33

Commented by ajfour last updated on 13/Aug/18

great, Tanmay Sir.

great,TanmaySir.

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