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Question Number 54480 by 951172235v last updated on 04/Feb/19

show that  tan α +tan (α+((2Λ^− )/5)) +tan (α+((4Λ^− )/5)) +tan (α+((6Λ^− )/5)) + tan (α+((8Λ^− )/5)) = 5tan 5α

showthattanα+tan(α+2Λ5)+tan(α+4Λ5)+tan(α+6Λ5)+tan(α+8Λ5)=5tan5α

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

tanα+tan(α+72^o )+tan(α+144^o )+tan(α+216^o )+tan(α+288^o )  tanα=a  tan72^o =cot18^o =(√(5+2(√5)))  =b  tan144^0 =tan(180−36)=−tan36^o =−(√(5−2(√5))) =−c  tan(216)=tan(180+36)=tan36=(√(5−2(√5) ))  =c  tan(288)=tan(3×90+18)=−cot18=−(√(5+2(√5))) =−b  (5+2(√5) )(5−2(√5) )=5  so let tanα=a  (√(5+2(√5) )) =b  (√(5−2(√5))) =c  bc=(√5)   tanα+((tanα+tan72)/(1−tanαtan72))+((tanα+tan144)/(1−tanαtan144))+((tanα+tan216)/(1−tanαtan216))+((tanα+tan288)/(1−tanαtan288))  a+((a+b)/(1−ab))+((a−c)/(1+ac))+((a+c)/(1−ac))+((a−b)/(1+ab))  a+(((a+b)(1+ab)+(a−b)(1−ab))/(1−a^2 b^2 ))+(((a+c)(1+ac)+(a−c)(1−ac))/(1−a^2 c^2 ))  a+((a+b+a^2 b+ab^2 +a−b−a^2 b+ab^2 )/(1−a^2 b^2 ))+((a+c+a^2 c+ac^2 +a−c−a^2 c+ac^2 )/(1−a^2 c^2 ))  a+((2a+2ab^2 )/(1−a^2 b^2 ))+((2a+2ac^2 )/(1−a^2 c^2 ))  a+((2a(1+ab))/((1+ab)(1−ab)))+((2a(1+ac))/((1+ac)(1−ac)))  a+((2a)/(1−ab))+((2a)/(1−ac))  a+2a(((1−ac+1−ab)/(1−ab−ac+a^2 bc)))  a+2a(((2−a(b+c))/(1−a(b+c)+a^2 bc)))  ((a−a^2 (b+c)+a^3 bc+4a−2a^2 (b+c))/(1−ab−ac+a^2 bc))  ((5a−3a^2 (b+c)+a^3 bc)/(1−ab−ac+a^2 bc))  wait...

tanα+tan(α+72o)+tan(α+144o)+tan(α+216o)+tan(α+288o)tanα=atan72o=cot18o=5+25=btan1440=tan(18036)=tan36o=525=ctan(216)=tan(180+36)=tan36=525=ctan(288)=tan(3×90+18)=cot18=5+25=b(5+25)(525)=5solettanα=a5+25=b525=cbc=5tanα+tanα+tan721tanαtan72+tanα+tan1441tanαtan144+tanα+tan2161tanαtan216+tanα+tan2881tanαtan288a+a+b1ab+ac1+ac+a+c1ac+ab1+aba+(a+b)(1+ab)+(ab)(1ab)1a2b2+(a+c)(1+ac)+(ac)(1ac)1a2c2a+a+b+a2b+ab2+aba2b+ab21a2b2+a+c+a2c+ac2+aca2c+ac21a2c2a+2a+2ab21a2b2+2a+2ac21a2c2a+2a(1+ab)(1+ab)(1ab)+2a(1+ac)(1+ac)(1ac)a+2a1ab+2a1aca+2a(1ac+1ab1abac+a2bc)a+2a(2a(b+c)1a(b+c)+a2bc)aa2(b+c)+a3bc+4a2a2(b+c)1abac+a2bc5a3a2(b+c)+a3bc1abac+a2bcwait...

Answered by 951172235v last updated on 05/Feb/19

tan (α+((8  Λ^− )/5)) =tan (2Λ^− −((2Λ^− )/5) +α) =tan (α−((2Λ^− )/5))  tan (α+((2Λ^− )/5))+tan (α+((8Λ^− )/5)) = ((x+y)/(1−xy))  + ((x−y)/(1+xy))  where tanα =x  tan((2Λ^− )/5) =y  = ((2x(1+y^2 ))/(1−x^2 y^2 ))  = ((2x((√5) +1)^2 )/(1−(5+2(√5))x^2 ))  similarly tan (α+((4Λ^− )/5))+tan (α+((6Λ^− )/5))  =  ((2x((√5) −1)^2 )/(1−(5−2(√5))x^2 ))  L.H.S = x+2x[((((√5) −1)^2 {1−(5+2(√5))x^2 +((√5) +1)^2 {1−(5−2(√5))x^2 })/(1−10x^2 +5x^4 ))                 =x+2x(((12−(5×12−8×5)x^2 )/(1−10x^2 +5x^4 )))         = ((5(5x−10x^3 +x^5 ))/(1−10x^2 +5x^4 ))         = 5tan 5α   ans.

tan(α+8Λ5)=tan(2Λ2Λ5+α)=tan(α2Λ5)tan(α+2Λ5)+tan(α+8Λ5)=x+y1xy+xy1+xywheretanα=xtan2Λ5=y=2x(1+y2)1x2y2=2x(5+1)21(5+25)x2similarlytan(α+4Λ5)+tan(α+6Λ5)=2x(51)21(525)x2L.H.S=x+2x[(51)2{1(5+25)x2+(5+1)2{1(525)x2}110x2+5x4=x+2x(12(5×128×5)x2110x2+5x4)=5(5x10x3+x5)110x2+5x4=5tan5αans.

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