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} $