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

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