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If $\theta=\frac{\pi}{9}$, then $1+27 \tan ^2 \theta-33 \tan ^4 \theta+\tan ^6 \theta=$
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Verified Answer
The correct answer is:
$4$
We know, at $\theta=\frac{\pi}{9}$
$\begin{aligned}
& \tan 3 \theta=\tan \frac{\pi}{3} \\
& \Rightarrow \frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}=\sqrt{3} \\
& \Rightarrow\left(3 \tan \theta-\tan ^3 \theta\right)^2=3\left(1-3 \tan ^2 \theta\right)^2 \\
& \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\
& =3\left(1+9 \tan ^4 \theta-6 \tan ^2 \theta\right) \\
& \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\
& =3+27 \tan ^4 \theta-18 \tan ^2 \theta \\
& \Rightarrow \tan ^6 \theta+27 \tan ^2 \theta-33 \tan ^4 \theta+1=4
\end{aligned}$
$\begin{aligned}
& \tan 3 \theta=\tan \frac{\pi}{3} \\
& \Rightarrow \frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}=\sqrt{3} \\
& \Rightarrow\left(3 \tan \theta-\tan ^3 \theta\right)^2=3\left(1-3 \tan ^2 \theta\right)^2 \\
& \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\
& =3\left(1+9 \tan ^4 \theta-6 \tan ^2 \theta\right) \\
& \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\
& =3+27 \tan ^4 \theta-18 \tan ^2 \theta \\
& \Rightarrow \tan ^6 \theta+27 \tan ^2 \theta-33 \tan ^4 \theta+1=4
\end{aligned}$
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