Question

If $k\left[{\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}\right]=921$, then the value of $k$ is

Answer 1

Given $k[ta{n}^6\frac{\pi }{9}-33ta{n}^4\frac{\pi }{9}+27ta{n}^2\frac{\pi }{9}]=921$ ....(1)
Since, $\sqrt{3}=\tan \frac{\pi }{3}=\tan 3\left(\frac{\pi }{9}\right)$
$\Rightarrow \sqrt{3}=\frac{3\tan \left(\frac{\pi }{9}\right)-{\tan }^3\left(\frac{\pi }{9}\right)}{1-3{\tan }^2\left(\frac{\pi }{9}\right)}$
$\Rightarrow \sqrt{3}(1-3{\tan }^2(\frac{\pi }{9}\left)\right)=\tan \left(\frac{\pi }{9}\right)(3-{\tan }^2(\frac{\pi }{9}\left)\right)$

On squaring

$\Rightarrow 3(1-3{\tan }^2(\frac{\pi }{9}){)}^2={\tan }^2(\frac{\pi }{9}\left)\right(3-{\tan }^2\left(\frac{\pi }{9}\right){)}^2$

$3+27ta{n}^4\left(\frac{\pi }{9}\right)-18ta{n}^2\left(\frac{\pi }{9}\right)=9ta{n}^2\left(\frac{\pi }{9}\right)+ta{n}^6\left(\frac{\pi }{9}\right)-6ta{n}^4\left(\frac{\pi }{9}\right)$

$\Rightarrow {\tan }^6\left(\frac{\pi }{9}\right)-33{\tan }^2\left(\frac{\pi }{9}\right)+27{\tan }^2\left(\frac{\pi }{9}\right)=3$
Put this value in (1), we get
$\Rightarrow 3k=921$
$\Rightarrow k=307$.