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Question Number 39388 by maxmathsup by imad last updated on 05/Jul/18

calculate A =tan((π/5)).tan(((2π)/5)).tan(((3π)/5)).tan(((4π)/5))

$${calculate}\:{A}\:={tan}\left(\frac{\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

tan36.tan72.tan108.tan144 tan108=tan(180−72)=−tan72 tan144=tan(180−36)=−tan36 tan^2 36×tan^2 72 (tan36.tan18)^2 ((((√(10−2(√5) ))/4)/(((√(5 )) +1)/4))×((((√5) −1)/4)/((√(10+2(√5) ))/4)))^2 =(((10−2(√(5 )))/(10+2(√5) ))×(((√5) −1)/((√5) +1)))^2 devide N_r andDr by 2(√5) =({((((√5) −1)/((√5) +1)))^2 }^2 =(((6−2(√5))/(6+2(√5))))^2 =(((3−(√5) )/(3+(√5))))^2 =(((14−6(√(5 )))/(14+6(√5))))

$${tan}\mathrm{36}.{tan}\mathrm{72}.{tan}\mathrm{108}.{tan}\mathrm{144} \\ $$$${tan}\mathrm{108}={tan}\left(\mathrm{180}−\mathrm{72}\right)=−{tan}\mathrm{72} \\ $$$${tan}\mathrm{144}={tan}\left(\mathrm{180}−\mathrm{36}\right)=−{tan}\mathrm{36} \\ $$$${tan}^{\mathrm{2}} \mathrm{36}×{tan}^{\mathrm{2}} \mathrm{72} \\ $$$$\left({tan}\mathrm{36}.{tan}\mathrm{18}\right)^{\mathrm{2}} \\ $$$$\left(\frac{\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}\:\:}}{\mathrm{4}}}{\frac{\sqrt{\mathrm{5}\:}\:\:+\mathrm{1}}{\mathrm{4}}}×\frac{\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{4}}}{\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}\:}}{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}\:}}{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}\:}×\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\sqrt{\mathrm{5}}\:+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$${devide}\:{N}_{{r}} {andDr}\:{by}\:\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$$=\left(\left\{\left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\sqrt{\mathrm{5}}\:+\mathrm{1}}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} =\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \:=\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}\:}{\mathrm{3}+\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \right. \\ $$$$=\left(\frac{\mathrm{14}−\mathrm{6}\sqrt{\mathrm{5}\:}}{\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}}}\right) \\ $$

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