Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Use tan 3x with x = 10 degrees, where tan 30 degrees is 1 over root 3, and square the identity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(t=\tan10^{\circ}\). Then \(\tan30^{\circ}=\dfrac{3t-t^{3}}{1-3t^{2}}=\dfrac{1}{\sqrt{3}}\).

Step 2: Square
\(3(3t-t^{3})^{2}=(1-3t^{2})^{2}\), so \(3t^{2}(9-6t^{2}+t^{4})=1-6t^{2}+9t^{4}\).

Step 3: Expand
\(27t^{2}-18t^{4}+3t^{6}=1-6t^{2}+9t^{4}\).

Step 4: Rearrange
\(3t^{6}-27t^{4}+33t^{2}=1\), option (B).

Final Answer:
The expression equals 1. \[ \boxed{\text{(B) }1} \]