What is \frac{2\sin^{3}\theta-\sin\theta}{\cos\theta-2\cos^{3}\theta} , (0^{\circ} \lt \theta \lt 90^{\circ}) equal to?
- A. \sin\theta
- B. \cos\theta
- C. \tan\theta ✓
- D. \cot\theta
Correct Answer: C. \tan\theta
Explanation
Factor the numerator and denominator: \frac{\sin\theta(2\sin^2\theta - 1)}{\cos\theta(1 - 2\cos^2\theta)}. Both (2\sin^2\theta - 1) and (1 - 2\cos^2\theta) equal -\cos 2\theta. Canceling these terms leaves \frac{\sin\theta}{\cos\theta} = \tan\theta.
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