<p>If \(\frac{\cos(\alpha+\gamma)}{\cos(\alpha-\gamma)} = \cos 2\beta\) then \(\tan\alpha\), \(\tan\beta\) and \(\tan\gamma\) are in</p>
Step-by-Step Solution
Key Concept: Apply the product-to-sum formula by expanding cos(α+γ) and cos(α-γ), then use the given ratio to establish a relationship between the tangent functions that reveals their progression.
<p><strong>Step 1:</strong> Expand using product-to-sum formulas:</p><p>cos(α+γ) = cos α cos γ - sin α sin γ</p><p>cos(α-γ) = cos α cos γ + sin α sin γ</p><p><strong>Step 2:</strong> Substitute into the given equation:</p><p>$$\frac{\cos α \cos γ - \sin α \sin γ}{\cos α \cos γ + \sin α \sin γ} = \cos 2β$$</p><p><strong>Step 3:</strong> Divide numerator and denominator by cos α cos γ:</p><p>$$\frac{1 - \tan α \tan γ}{1 + \tan α \tan γ} = \cos 2β$$</p><p><strong>Step 4:</strong> Recognize that the left side equals cot(α+γ). Also, using the double angle formula:</p><p>$$\frac{1 - \tan α \tan γ}{1 + \tan α \tan γ} = \frac{1 - \tan^2 β}{1 + \tan^2 β}$$</p><p><strong>Step 5:</strong> This gives us: 1 - tan α tan γ = 1 - tan² β and 1 + tan α tan γ = 1 + tan² β</p><p>Therefore: <strong>tan α tan γ = tan² β</strong></p><p><strong>Step 6:</strong> Taking square root: tan β = √(tan α tan γ)</p><p>This is the geometric mean relation, meaning tan α, tan β, tan γ are in <strong>Geometric Progression</strong></p><p>∴ Answer: B</p>
Correct Answer: B