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

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

Answered by math1967 last updated on 31/Jan/19

$$let\frac{\pi }{9}=\theta \therefore tan3\theta =\sqrt{3}$$$$\Rightarrow \frac{3tan\theta -\tan {}^3\theta }{1-3\tan {}^2\theta }=\sqrt{3}$$$$\Rightarrow {(3\tan \theta -\tan {}^3\theta )}^2=3{(1-3{tan}^2\theta )}^2$$$$\Rightarrow 9\tan {}^2\theta -6\tan {}^4\theta +\tan {}^6\theta =3-18\tan {}^2\theta +27\tan {}^4\theta$$$$27\tan {}^2\theta -33\tan {}^4\theta +\tan {}^6\theta =3$$$$\therefore \tan {}^6\frac{\pi }{9}-33\tan {}^4\frac{\pi }{9}+27\tan {}^2\frac{\pi }{9}=3ans$$

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