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


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

$show that$ $\tan α + \tan (α + \frac{2 \bar{Λ}}{5}) + \tan (α + \frac{4 \bar{Λ}}{5}) + \tan (α + \frac{6 \bar{Λ}}{5}) + \tan (α + \frac{8 \bar{Λ}}{5}) = 5 \tan 5 α$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

	$t a n α + t a n (α + 72^{o}) + t a n (α + 144^{o}) + t a n (α + 216^{o}) + t a n (α + 288^{o})$ $t a n α = a$ $t a n 72^{o} = c o t 18^{o} = \sqrt{5 + 2 \sqrt{5}} = b$ $t a n 144^{0} = t a n (180 - 36) = - t a n 36^{o} = - \sqrt{5 - 2 \sqrt{5}} = - c$ $t a n (216) = t a n (180 + 36) = t a n 36 = \sqrt{5 - 2 \sqrt{5}} = c$ $t a n (288) = t a n (3 \times 90 + 18) = - c o t 18 = - \sqrt{5 + 2 \sqrt{5}} = - b$ $(5 + 2 \sqrt{5}) (5 - 2 \sqrt{5}) = 5$ $s o l e t t a n α = a$ $\sqrt{5 + 2 \sqrt{5}} = b$ $\sqrt{5 - 2 \sqrt{5}} = c$ $b c = \sqrt{5}$ $t a n α + \frac{t a n α + t a n 72}{1 - t a n α t a n 72} + \frac{t a n α + t a n 144}{1 - t a n α t a n 144} + \frac{t a n α + t a n 216}{1 - t a n α t a n 216} + \frac{t a n α + t a n 288}{1 - t a n α t a n 288}$ $a + \frac{a + b}{1 - a b} + \frac{a - c}{1 + a c} + \frac{a + c}{1 - a c} + \frac{a - b}{1 + a b}$ $a + \frac{(a + b) (1 + a b) + (a - b) (1 - a b)}{1 - a^{2} b^{2}} + \frac{(a + c) (1 + a c) + (a - c) (1 - a c)}{1 - a^{2} c^{2}}$ $a + \frac{a + b + a^{2} b + {a b}^{2} + a - b - a^{2} b + {a b}^{2}}{1 - a^{2} b^{2}} + \frac{a + c + a^{2} c + {a c}^{2} + a - c - a^{2} c + {a c}^{2}}{1 - a^{2} c^{2}}$ $a + \frac{2 a + 2 {a b}^{2}}{1 - a^{2} b^{2}} + \frac{2 a + 2 {a c}^{2}}{1 - a^{2} c^{2}}$ $a + \frac{2 a (1 + a b)}{(1 + a b) (1 - a b)} + \frac{2 a (1 + a c)}{(1 + a c) (1 - a c)}$ $a + \frac{2 a}{1 - a b} + \frac{2 a}{1 - a c}$ $a + 2 a (\frac{1 - a c + 1 - a b}{1 - a b - a c + a^{2} b c})$ $a + 2 a (\frac{2 - a (b + c)}{1 - a (b + c) + a^{2} b c})$ $\frac{a - a^{2} (b + c) + a^{3} b c + 4 a - 2 a^{2} (b + c)}{1 - a b - a c + a^{2} b c}$ $\frac{5 a - 3 a^{2} (b + c) + a^{3} b c}{1 - a b - a c + a^{2} b c}$ $w a i t . . .$
	Answered by 951172235v last updated on 05/Feb/19
	$tan (α+((8 Λ^− )/5)) =tan (2Λ^− −((2Λ^− )/5) +α) =tan (α−((2Λ^− )/5)) tan (α+((2Λ^− )/5))+tan (α+((8Λ^− )/5)) = ((x+y)/(1−xy)) + ((x−y)/(1+xy)) where tanα =x tan((2Λ^− )/5) =y = ((2x(1+y^2 ))/(1−x^2 y^2 )) = ((2x((√5) +1)^2 )/(1−(5+2(√5))x^2 )) similarly tan (α+((4Λ^− )/5))+tan (α+((6Λ^− )/5)) = ((2x((√5) −1)^2 )/(1−(5−2(√5))x^2 )) L.H.S = x+2x[((((√5) −1)^2 {1−(5+2(√5))x^2 +((√5) +1)^2 {1−(5−2(√5))x^2 })/(1−10x^2 +5x^4 )) =x+2x(((12−(5×12−8×5)x^2 )/(1−10x^2 +5x^4 ))) = ((5(5x−10x^3 +x^5 ))/(1−10x^2 +5x^4 )) = 5tan 5α ans.$
	$\tan (α + \frac{8 \bar{Λ}}{5}) = \tan (2 \bar{Λ} - \frac{2 \bar{Λ}}{5} + α) = \tan (α - \frac{2 \bar{Λ}}{5})$ $\tan (α + \frac{2 \bar{Λ}}{5}) + \tan (α + \frac{8 \bar{Λ}}{5}) = \frac{x + y}{1 - xy} + \frac{x - y}{1 + xy}$ $where \tan α = x \tan \frac{2 \bar{Λ}}{5} = y$ $= \frac{2 x (1 + y^{2})}{1 - x^{2} y^{2}} = \frac{2 x {(\sqrt{5} + 1)}^{2}}{1 - (5 + 2 \sqrt{5}) x^{2}}$ $similarly \tan (α + \frac{4 \bar{Λ}}{5}) + \tan (α + \frac{6 \bar{Λ}}{5})$ $= \frac{2 x {(\sqrt{5} - 1)}^{2}}{1 - (5 - 2 \sqrt{5}) x^{2}}$ $L . H . S = x + 2 x [\frac{{(\sqrt{5} - 1)}^{2} {1 - (5 + 2 \sqrt{5}) x^{2} + {(\sqrt{5} + 1)}^{2} {1 - (5 - 2 \sqrt{5}) x^{2}}}{1 - {10 x}^{2} + {5 x}^{4}}$ $= x + 2 x (\frac{12 - (5 \times 12 - 8 \times 5) x^{2}}{1 - {10 x}^{2} + {5 x}^{4}})$ $= \frac{5 (5 x - {10 x}^{3} + x^{5})}{1 - {10 x}^{2} + {5 x}^{4}}$ $= 5 \tan 5 α ans .$
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