Question

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

• A. 0
• B. $\sqrt{3}$
• C. 3
• D. 9

Answer 1

The correct option is A 0

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

let $\frac{\pi }{9}=\theta$

$\tan 3\theta =\tan \left(\frac{\pi }{3}\right)$

$\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\sqrt{3}$

${(3\tan \theta -{\tan }^3\theta )}^2=3{(1-3{\tan }^2\theta )}^2$

$9{\tan }^2\theta +{\tan }^6\theta -6{\tan }^4\theta =3+27{\tan }^4\theta -18{\tan }^2\theta$

${\tan }^6\theta -33{\tan }^4\theta +27{\tan }^2\theta =0$

${\tan }^{-6}\frac{1}{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}=0$

option a is correct.