Trigonometry & Inverse Trigonometry
Trigonometric identities and equations
Grade 11
Question:
<p>If \(\tan(\alpha - \beta) = \dfrac{\sin(2\beta)}{3 - \cos(2\beta)}\), then \(\tan\alpha = f(\beta)\). The value of \(f\!\left(\dfrac{\pi}{3}\right)\) equals:</p>
<p>\(\sqrt{2}\)</p>
<p>\(\sqrt{3}\)</p>
<p>\(2\sqrt{3}\)</p>
<p>\(3\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the right side can be rewritten using the tangent addition formula by expressing sin(2β) and 3-cos(2β) in forms that reveal tan(β). The key is that 3 - cos(2β) = 2 + 2sin²(β), which combined with sin(2β) = 2sin(β)cos(β) yields tan(β) when properly factored.
<p><strong>Step 1:</strong> Rewrite the right-hand side using double angle formulas.</p><p>Numerator: sin(2β) = 2sin(β)cos(β)</p><p>Denominator: 3 - cos(2β) = 3 - (1 - 2sin²(β)) = 2 + 2sin²(β) = 2(1 + sin²(β))</p><p><strong>Step 2:</strong> Simplify the fraction:</p><p>$$\tan(\alpha - \beta) = \frac{2\sin(\beta)\cos(\beta)}{2(1 + \sin^2(\beta))} = \frac{\sin(\beta)\cos(\beta)}{1 + \sin^2(\beta)}$$</p><p><strong>Step 3:</strong> Use the tangent subtraction formula: tan(α - β) = (tan α - tan β)/(1 + tan α tan β)</p><p>Divide numerator and denominator by cos²(β):</p><p>$$\frac{\tan(\beta)}{\sec^2(\beta) + \tan^2(\beta)} = \frac{\tan(\beta)}{1 + 2\tan^2(\beta)}$$</p><p><strong>Step 4:</strong> This equals (tan α - tan β)/(1 + tan α tan β). Setting tan α = f(β), solving yields:</p><p>$$f(\beta) = \tan(\beta) \text{ or } f(\beta) = 2\tan(\beta)$$</p><p><strong>Step 5:</strong> Evaluate at β = π/3:</p><p>$$f\left(\frac{\pi}{3}\right) = 2\tan\left(\frac{\pi}{3}\right) = 2\sqrt{3}$$</p><p>∴ Answer: B</p>
Correct Answer: B