Question

The value of ${\tan }^{6}20^{\circ }-33{\tan }^{4}20^{\circ }+27{\tan }^{2}20^{\circ }$ is :

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

$\sqrt{3}=\tan 60^{\circ }=\tan \left(3\times 20^{\circ }\right)=\frac{3\tan 20^{\circ }-{\tan }^{3}20^{\circ }}{1-3{\tan }^{2}20^{\circ }}$
Squaring,
$3=\frac{9t^2+t^6-6t^4}{1+9t^4-6t^2},\tan 20^{\circ }=t$
$\Rightarrow t^6-33t^4+27t^2=3$
$\Rightarrow {\tan }^{6}20^{\circ }-33{\tan }^{4}20^{\circ }+27{\tan }^{2}20^{\circ }=3$