Question:

Step 1: Simplify the right-hand side using double angle formulas. We start with the given equation: $$\tan(\alpha - \beta) = \frac{\sin(2\beta)}{3 - \cos(2\beta)}$$ Let's simplify the denominator using the double angle formula $\cos(2\beta) = 1 - 2\sin^2\beta$: $$3 - \cos(2\beta) = 3 - (1 - 2\sin^2\beta) = 2 + 2\sin^2\beta = 2(1 + \sin^2\beta)$$ For the numerator, we use $\sin(2\beta) = 2\sin\beta\cos\beta$: $$\frac{\sin(2\beta)}{3 - \cos(2\beta)} = \frac{2\sin\beta\cos\beta}{2(1 + \sin^2\beta)} = \frac{\sin\beta\cos\beta}{1 + \sin^2\beta}$$ Step 2: Convert the simplified expression to a form involving $\tan\beta$. Divide both numerator and denominator by $\cos^2\beta$: $$\frac{\sin\beta\cos\beta}{1 + \sin^2\beta} = \frac{\frac{\sin\beta\cos\beta}{\cos^2\beta}}{\frac{1 + \sin^2\beta}{\cos^2\beta}} = \frac{\tan\beta}{\sec^2\beta + \tan^2\beta}$$ Since $\sec^2\beta = 1 + \tan^2\beta$, we have: $$\sec^2\beta + \tan^2\beta = 1 + \tan^2\beta + \tan^2\beta = 1 + 2\tan^2\beta$$ Therefore: $$\tan(\alpha - \beta) = \frac{\tan\beta}{1 + 2\tan^2\beta}$$ Step 3: Conjecture that $\tan\alpha = 2\tan\beta$ and verify. Let's test whether $\tan\alpha = 2\tan\beta$. Using the tangent subtraction formula: $$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \frac{2\tan\beta - \tan\beta}{1 + 2\tan\beta \cdot \tan\beta} = \frac{\tan\beta}{1 + 2\tan^2\beta}$$ This matches our simplified expression from Step 2, confirming that: $$f(\beta) = \tan\alpha = 2\tan\beta$$ Step 4: Evaluate $f\left(\dfrac{\pi}{3}\right)$. Now we substitute $\beta = \dfrac{\pi}{3}$: $$f\left(\frac{\pi}{3}\right) = 2\tan\left(\frac{\pi}{3}\right) = 2 \cdot \sqrt{3} = 2\sqrt{3}$$ **Final Answer:** The value of $f\left(\dfrac{\pi}{3}\right) = 2\sqrt{3}$, which corresponds to **Option 2**.