tan-6-pi-9-33-tan-4-pi-9-27tan-2-pi-9-

Question Number 54219 by 951172235v last updated on 31/Jan/19

\tan^{6} \frac{π}{9} - 33 \tan^{4} \frac{π}{9} + {27 \tan}^{2} \frac{π}{9} =

Answered by math1967 last updated on 31/Jan/19

l e t \frac{π}{9} = θ ∴ t a n 3 θ = \sqrt{3}

\Rightarrow \frac{3 t a n θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} = \sqrt{3}

\Rightarrow {(3 \tan θ - \tan^{3} θ)}^{2} = 3 {(1 - 3 {t a n}^{2} θ)}^{2}

\Rightarrow 9 \tan^{2} θ - 6 \tan^{4} θ + \tan^{6} θ = 3 - 18 \tan^{2} θ + 27 \tan^{4} θ

27 \tan^{2} θ - 33 \tan^{4} θ + \tan^{6} θ = 3

∴ \tan^{6} \frac{π}{9} - 33 \tan^{4} \frac{π}{9} + 27 \tan^{2} \frac{π}{9} = 3 a n s

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