Question

If, ${\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^2\frac{\pi }{9}=p$, then the value of p =

Answer 1

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

$\frac{3\tan \frac{\pi }{9}-{\tan }^3\frac{\pi }{9}}{1-3{\tan }^2\frac{\pi }{2}}=\tan \frac{\pi }{3}=\sqrt{3}$

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

squaring both sides

${\tan }^6\frac{\pi }{9}-6{\tan }^4\frac{\pi }{9}+{\tan }^2\frac{\pi }{9}=3(3{\tan }^4\frac{\pi }{9}-6{\tan }^2\frac{\pi }{9}+1)$

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

Rearrange to get

${\tan }^6\frac{\pi }{9}-33{\tan }^4\frac{\pi }{9}+27{\tan }^4\left(\frac{\pi }{9}\right)=3$