<p>If \(\tan\theta + \tan\left(\frac{\pi}{3} + \theta\right) + \tan\left(\frac{2\pi}{3} + \theta\right) = k\tan 3\theta\) then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula and the identity that tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) when A + B + C = π. Here, the three angles sum to π, so their tangents satisfy a special multiplicative relationship that connects to tan(3θ).
<p><strong>Step 1:</strong> Recognize that θ + (π/3 + θ) + (2π/3 + θ) = π + 3θ, so the three angle arguments sum to π (modulo the 3θ term).</p><p><strong>Step 2:</strong> Note that if A + B + C = π, then tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C). However, here we need to use the tangent addition formula:</p><p>tan(π/3 + θ) = (tan(π/3) + tan(θ))/(1 - tan(π/3)tan(θ)) = (√3 + tan(θ))/(1 - √3tan(θ))</p><p>tan(2π/3 + θ) = (tan(2π/3) + tan(θ))/(1 - tan(2π/3)tan(θ)) = (-√3 + tan(θ))/(1 + √3tan(θ))</p><p><strong>Step 3:</strong> Let t = tan(θ). Adding the three terms and simplifying (combining fractions and factoring):</p><p>tan(θ) + tan(π/3 + θ) + tan(2π/3 + θ) = 3tan(θ) - tan³(θ)/(1 - 3tan²(θ)) = 3tan(θ) - tan³(θ) · 1/(1 - 3tan²(θ))</p><p><strong>Step 4:</strong> Recognize that tan(3θ) = (3tan(θ) - tan³(θ))/(1 - 3tan²(θ)). Therefore:</p><p>tan(θ) + tan(π/3 + θ) + tan(2π/3 + θ) = 3tan(3θ)</p><p>∴ k = <strong>3</strong></p>
Correct Answer: C