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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

$$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}\:\alpha\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{4}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{6}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\:\mathrm{tan}\:\left(\alpha+\frac{\mathrm{8}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:=\:\mathrm{5tan}\:\mathrm{5}\alpha \\ $$

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...

$${tan}\alpha+{tan}\left(\alpha+\mathrm{72}^{{o}} \right)+{tan}\left(\alpha+\mathrm{144}^{{o}} \right)+{tan}\left(\alpha+\mathrm{216}^{{o}} \right)+{tan}\left(\alpha+\mathrm{288}^{{o}} \right) \\ $$$${tan}\alpha={a} \\ $$$${tan}\mathrm{72}^{{o}} ={cot}\mathrm{18}^{{o}} =\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}\:\:={b} \\ $$$${tan}\mathrm{144}^{\mathrm{0}} ={tan}\left(\mathrm{180}−\mathrm{36}\right)=−{tan}\mathrm{36}^{{o}} =−\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\:=−{c} \\ $$$${tan}\left(\mathrm{216}\right)={tan}\left(\mathrm{180}+\mathrm{36}\right)={tan}\mathrm{36}=\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:}\:\:={c} \\ $$$${tan}\left(\mathrm{288}\right)={tan}\left(\mathrm{3}×\mathrm{90}+\mathrm{18}\right)=−{cot}\mathrm{18}=−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}\:=−{b} \\ $$$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)=\mathrm{5} \\ $$$${so}\:{let}\:{tan}\alpha={a} \\ $$$$\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:}\:={b} \\ $$$$\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\:={c} \\ $$$${bc}=\sqrt{\mathrm{5}}\: \\ $$$${tan}\alpha+\frac{{tan}\alpha+{tan}\mathrm{72}}{\mathrm{1}−{tan}\alpha{tan}\mathrm{72}}+\frac{{tan}\alpha+{tan}\mathrm{144}}{\mathrm{1}−{tan}\alpha{tan}\mathrm{144}}+\frac{{tan}\alpha+{tan}\mathrm{216}}{\mathrm{1}−{tan}\alpha{tan}\mathrm{216}}+\frac{{tan}\alpha+{tan}\mathrm{288}}{\mathrm{1}−{tan}\alpha{tan}\mathrm{288}} \\ $$$${a}+\frac{{a}+{b}}{\mathrm{1}−{ab}}+\frac{{a}−{c}}{\mathrm{1}+{ac}}+\frac{{a}+{c}}{\mathrm{1}−{ac}}+\frac{{a}−{b}}{\mathrm{1}+{ab}} \\ $$$${a}+\frac{\left({a}+{b}\right)\left(\mathrm{1}+{ab}\right)+\left({a}−{b}\right)\left(\mathrm{1}−{ab}\right)}{\mathrm{1}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} }+\frac{\left({a}+{c}\right)\left(\mathrm{1}+{ac}\right)+\left({a}−{c}\right)\left(\mathrm{1}−{ac}\right)}{\mathrm{1}−{a}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$${a}+\frac{{a}+{b}+{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} +{a}−{b}−{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} }+\frac{{a}+{c}+{a}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} +{a}−{c}−{a}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$${a}+\frac{\mathrm{2}{a}+\mathrm{2}{ab}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} }+\frac{\mathrm{2}{a}+\mathrm{2}{ac}^{\mathrm{2}} }{\mathrm{1}−{a}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$${a}+\frac{\mathrm{2}{a}\left(\mathrm{1}+{ab}\right)}{\left(\mathrm{1}+{ab}\right)\left(\mathrm{1}−{ab}\right)}+\frac{\mathrm{2}{a}\left(\mathrm{1}+{ac}\right)}{\left(\mathrm{1}+{ac}\right)\left(\mathrm{1}−{ac}\right)} \\ $$$${a}+\frac{\mathrm{2}{a}}{\mathrm{1}−{ab}}+\frac{\mathrm{2}{a}}{\mathrm{1}−{ac}} \\ $$$${a}+\mathrm{2}{a}\left(\frac{\mathrm{1}−{ac}+\mathrm{1}−{ab}}{\mathrm{1}−{ab}−{ac}+{a}^{\mathrm{2}} {bc}}\right) \\ $$$${a}+\mathrm{2}{a}\left(\frac{\mathrm{2}−{a}\left({b}+{c}\right)}{\mathrm{1}−{a}\left({b}+{c}\right)+{a}^{\mathrm{2}} {bc}}\right) \\ $$$$\frac{{a}−{a}^{\mathrm{2}} \left({b}+{c}\right)+{a}^{\mathrm{3}} {bc}+\mathrm{4}{a}−\mathrm{2}{a}^{\mathrm{2}} \left({b}+{c}\right)}{\mathrm{1}−{ab}−{ac}+{a}^{\mathrm{2}} {bc}} \\ $$$$\frac{\mathrm{5}{a}−\mathrm{3}{a}^{\mathrm{2}} \left({b}+{c}\right)+{a}^{\mathrm{3}} {bc}}{\mathrm{1}−{ab}−{ac}+{a}^{\mathrm{2}} {bc}} \\ $$$${wait}... \\ $$

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.

$$\mathrm{tan}\:\left(\alpha+\frac{\mathrm{8}\:\:\overset{−} {\Lambda}}{\mathrm{5}}\right)\:=\mathrm{tan}\:\left(\mathrm{2}\overset{−} {\Lambda}−\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\:+\alpha\right)\:=\mathrm{tan}\:\left(\alpha−\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\right) \\ $$$$\mathrm{tan}\:\left(\alpha+\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\right)+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{8}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:=\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{1}−\mathrm{xy}}\:\:+\:\frac{\mathrm{x}−\mathrm{y}}{\mathrm{1}+\mathrm{xy}} \\ $$$$\mathrm{where}\:\mathrm{tan}\alpha\:=\mathrm{x}\:\:\mathrm{tan}\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\:=\mathrm{y} \\ $$$$=\:\frac{\mathrm{2x}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} }\:\:=\:\frac{\mathrm{2x}\left(\sqrt{\mathrm{5}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right)\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{similarly}\:\mathrm{tan}\:\left(\alpha+\frac{\mathrm{4}\overset{−} {\Lambda}}{\mathrm{5}}\right)+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{6}\overset{−} {\Lambda}}{\mathrm{5}}\right) \\ $$$$=\:\:\frac{\mathrm{2x}\left(\sqrt{\mathrm{5}}\:−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}−\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\right)\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{L}.\mathrm{H}.\mathrm{S}\:=\:\mathrm{x}+\mathrm{2x}\left[\frac{\left(\sqrt{\mathrm{5}}\:−\mathrm{1}\right)^{\mathrm{2}} \left\{\mathrm{1}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right)\mathrm{x}^{\mathrm{2}} +\left(\sqrt{\mathrm{5}}\:+\mathrm{1}\right)^{\mathrm{2}} \left\{\mathrm{1}−\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\right)\mathrm{x}^{\mathrm{2}} \right\}\right.}{\mathrm{1}−\mathrm{10x}^{\mathrm{2}} +\mathrm{5x}^{\mathrm{4}} }\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}+\mathrm{2x}\left(\frac{\mathrm{12}−\left(\mathrm{5}×\mathrm{12}−\mathrm{8}×\mathrm{5}\right)\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{10x}^{\mathrm{2}} +\mathrm{5x}^{\mathrm{4}} }\right) \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{5}\left(\mathrm{5x}−\mathrm{10x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{5}} \right)}{\mathrm{1}−\mathrm{10x}^{\mathrm{2}} +\mathrm{5x}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{5tan}\:\mathrm{5}\alpha\:\:\:\mathrm{ans}. \\ $$

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