tan^2 20+tan^2 40+tan^2 80=33 (please solve this and i want to know that which standard that question belongs?) help needed


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 54537 by hsbebeb last updated on 05/Feb/19

${t a n}^{2} 20 + {t a n}^{2} 40 + {t a n}^{2} 80 = 33$ $(p l e a s e s o l v e t h i s a n d i w a n t t o k n o w$ $t h a t w h i c h s t a n d a r d t h a t q u e s t i o n b e l o n g s ?)$ $h e l p n e e d e d$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

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

$d o w n l o a d t h i s b o o k f r e e f r o m w e b s i t e$ $a r c h i v e . o r g$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

	$θ = 20^{o}$ $9 θ = π$ $t a n 9 θ = t a n π = 0$ $f o r m u l a t a n (n θ) = \frac{s_{1} - s_{3} + s_{5} - s_{7} + . . .}{1 - s_{2} + s_{4} - . . .}$ $t a n 9 θ = 0 = \frac{9 c_{1} t a n θ - 9 c_{3} {t a n}^{3} θ + 9 c_{5} {t a n}^{5} θ - 9 c_{7} {t a n}^{7} θ + 9 c_{9} {t a n}^{9} θ}{D_{r}}$ $t a n θ = x$ $9 x - \frac{9 \times 8 \times 7}{3 \times 2} x^{3} + \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} x^{5} - \frac{9 \times 8}{2} x^{7} + x^{9} = 0$ $9 - 84 x^{2} + 14 \times 9 x^{4} - 36 x^{6} + x^{8} = 0$ $r o o t s o f t h i s e q n a r e t a n θ, t a n 2 θ, . . . [θ = 20^{o} . . .]$ $y = x^{2}$ $y^{4} - 36 y^{2} + 126 y^{2} - 84 y + 9 = 0$ $r o o t s o f t h i s e q n {t a n}^{2} θ, {t a n}^{2} 2 θ, {t a n}^{2} 3 θ, {t a n}^{2} 4 θ$ $s u m o f r o o t s = - (- 36) = 36$ ${t a n}^{2} 20 + {t a n}^{2} 40 + {t a n}^{2} 60 + {t a n}^{2} 80 = 36$ $s o {t a n}^{2} 20 + {t a n}^{2} 40 + {t a n}^{2} 80 = 36 - {(\sqrt{3})}^{2} = 33$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum