MHT CET · Maths · Trigonometric Ratios & Identities
\(3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=\)
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
Let \(\theta = 10^{\circ}\). We use the formula \(\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\). For \(\theta = 10^{\circ}\), \(\tan(30^{\circ}) = \frac{3\tan 10^{\circ} - \tan^3 10^{\circ}}{1 - 3\tan^2 10^{\circ}}\).
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