Question

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

Answer 1

${\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}$
Assuming $\tan \frac{\pi }{9}=t$
${\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}=t^6-33t^4+27t^2\cdots \left(1\right)$

Since, $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$
Putting $\theta =\frac{\pi }{9}$, we get
$\tan \frac{\pi }{3}=\frac{3\tan \frac{\pi }{9}-{\tan }^3\frac{\pi }{9}}{1-3{\tan }^2\frac{\pi }{9}}\Rightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\Rightarrow \sqrt{3}=\frac{t(3-t^2)}{1-3t^2}\Rightarrow \sqrt{3}(1-3t^2)=t(3-t^2)$
Squaring on both sides,
$3(1+9t^4-6t^2)=t^2(9+t^4-6t^2)\Rightarrow 3=t^6-33t^4+27t^2$
From equation $\left(1\right)$,
${\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}=3$