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Q.
The value of $\tan ^{6} 20^{\circ}-33\,\tan ^{4} 20^{\circ}+27 \,\tan ^{2} 20^{\circ}$ is :
Trigonometric Functions
Solution:
$\sqrt{3}=\tan 60^{\circ}=\tan \left(3 \times 20^{\circ}\right)=\frac{3 \tan 20^{\circ}-\tan ^{3} 20^{\circ}}{1-3 \tan ^{2} 20^{\circ}}$
Squaring,
$3=\frac{9 t ^{2}+ t ^{6}-6 t ^{4}}{1+9 t ^{4}-6 t ^{2}}, \tan 20^{\circ}= t $
$\Rightarrow t ^{6}-33 t ^{4}+27 t ^{2}=3$
$\Rightarrow \tan ^{6} 20^{\circ}-33 \tan ^{4} 20^{\circ}+27 \tan ^{2} 20^{\circ}=3$