tan^6 (π/9)−33tan^4 (π/9)+27tan^2 (π/9)=?


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Question Number 17420 by sushmitak last updated on 05/Jul/17

$\tan^{6} \frac{π}{9} - {33 \tan}^{4} \frac{π}{9} + {27 \tan}^{2} \frac{π}{9} = ?$
	Answered by Tinkutara last updated on 05/Jul/17

	$\tan 3 (\frac{π}{9}) = \sqrt{3}$ $\sqrt{3} = \frac{3 \tan \frac{π}{9} - \tan^{3} \frac{π}{9}}{1 - 3 \tan^{2} \frac{π}{9}}$ $3 (1 + 9 \tan^{4} \frac{π}{9} - 6 \tan^{2} \frac{π}{9}) =$ $\tan^{2} \frac{π}{9} (9 + \tan^{4} \frac{π}{9} - 6 \tan^{2} \frac{π}{9})$ $On rearranging, we get$ $\tan^{6} \frac{π}{9} - 33 \tan^{4} \frac{π}{9} + 27 \tan^{2} \frac{π}{9} = 3$
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