Question

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

• A. 9
• B. 0
• C. $\sqrt{3}$
• D. 3

Answer 1

Answer: (D)

We know that $\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$
Substituting $A=\frac{\pi }{9}$ and squaring, we get
${\left(\sqrt{3}\right)}^2={\left\{\frac{3\tan \frac{\pi }{9}-{\tan }^3\frac{\pi }{9}}{1-3{\tan }^2\frac{\pi }{9}}\right\}}^2$
$\Rightarrow 3-18{\tan }^2\frac{\pi }{9}+27{\tan }^4\frac{\pi }{9}$
$=9{\tan }^2\frac{\pi }{9}+{\tan }^6\frac{\pi }{9}-6{\tan }^4\frac{\pi }{9}$