3 ^6 10^ -27 ^4 10^ +33 ^2 10^ =1 - Vidyadip

Question

$3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=1$

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Answer

Since, $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\begin{array}{ll}
\therefore \quad \tan 3\left(10^{\circ}\right)=\frac{1}{\sqrt{3}}\\
\therefore \quad & \frac{3 \tan 10^{\circ}-\tan ^3 10^{\circ}}{1-3 \tan ^2 10^{\circ}}=\frac{1}{\sqrt{3}}
\end{array}$
Squaring both sides, we get
$\begin{array}{ll}
& \frac{\left(3 \tan 10^{\circ}-\tan ^3 10^{\circ}\right)^2}{\left(1-3 \tan ^2 10^{\circ}\right)^2}=\frac{1}{3} \\
\therefore \quad & \frac{9 \tan ^2 10^{\circ}-6 \tan ^4 10^{\circ}+\tan ^6 10^{\circ}}{1-6 \tan ^2 10^{\circ}+9 \tan ^4 10^{\circ}}=\frac{1}{3} \\
\therefore \quad & 3\left(9 \tan ^2 10^{\circ}-6 \tan ^4 10^{\circ}+\tan ^6 10^{\circ}\right) \\
=1-6 \tan ^2 10^{\circ}+9 \tan ^4 10^{\circ} \\
\therefore \quad & 27 \tan ^2 10^{\circ}-18 \tan ^4 10^{\circ}+3 \tan ^6 10^{\circ} \\
=1-6 \tan ^2 10^{\circ}+9 \tan { }^4 10^{\circ} \\
\therefore \quad & 3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=1
\end{array}$

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