If p \tan(\theta-30^{\circ}) = q \tan(\theta+120^{\circ}), then what is (p+q)/(p-q) equal to?
- A. \sin 2\theta
- B. \cos 2\theta
- C. 2 \sin 2\theta
- D. 2 \cos 2\theta ✓
Correct Answer: D. 2 \cos 2\theta
Explanation
From the given equation, \frac{p}{q} = \frac{\tan(\theta+120^{\circ})}{\tan(\theta-30^{\circ})}. Applying componendo and dividendo, \frac{p+q}{p-q} = \frac{\tan(\theta+120^{\circ}) + \tan(\theta-30^{\circ})}{\tan(\theta+120^{\circ}) - \tan(\theta-30^{\circ})}. Converting to sine and cosine, this simplifies to \frac{\sin(\theta+120^{\circ} + \theta-30^{\circ})}{\sin(\theta+120^{\circ} - (\theta-30^{\circ}))} = \frac{\sin(2\theta+90^{\circ})}{\sin 150^{\circ}} = \frac{\cos 2\theta}{1/2} = 2\cos 2\theta.