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


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

$c a l c u l a t e A = t a n (\frac{π}{5}) . t a n (\frac{2 π}{5}) . t a n (\frac{3 π}{5}) . t a n (\frac{4 π}{5})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18
	$tan36.tan72.tan108.tan144 tan108=tan(180−72)=−tan72 tan144=tan(180−36)=−tan36 tan^2 36×tan^2 72 (tan36.tan18)^2 ((((√(10−2(√5) ))/4)/(((√(5 )) +1)/4))×((((√5) −1)/4)/((√(10+2(√5) ))/4)))^2 =(((10−2(√(5 )))/(10+2(√5) ))×(((√5) −1)/((√5) +1)))^2 devide N_r andDr by 2(√5) =({((((√5) −1)/((√5) +1)))^2 }^2 =(((6−2(√5))/(6+2(√5))))^2 =(((3−(√5) )/(3+(√5))))^2 =(((14−6(√(5 )))/(14+6(√5))))$
	$t a n 36 . t a n 72 . t a n 108 . t a n 144$ $t a n 108 = t a n (180 - 72) = - t a n 72$ $t a n 144 = t a n (180 - 36) = - t a n 36$ ${t a n}^{2} 36 \times {t a n}^{2} 72$ ${(t a n 36 . t a n 18)}^{2}$ ${(\frac{\frac{\sqrt{10 - 2 \sqrt{5}}}{4}}{\frac{\sqrt{5} + 1}{4}} \times \frac{\frac{\sqrt{5} - 1}{4}}{\frac{\sqrt{10 + 2 \sqrt{5}}}{4}})}^{2}$ $= {(\frac{10 - 2 \sqrt{5}}{10 + 2 \sqrt{5}} \times \frac{\sqrt{5} - 1}{\sqrt{5} + 1})}^{2}$ $d e v i d e N_{r} a n d D r b y 2 \sqrt{5}$ $= ({{(\frac{\sqrt{5} - 1}{\sqrt{5} + 1})}^{2}}^{2} = {(\frac{6 - 2 \sqrt{5}}{6 + 2 \sqrt{5}})}^{2} = {(\frac{3 - \sqrt{5}}{3 + \sqrt{5}})}^{2}$ $= (\frac{14 - 6 \sqrt{5}}{14 + 6 \sqrt{5}})$
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