What is \frac{\sqrt{3}\cos 10^{\circ}-\sin 10^{\circ}}{\sin 25^{\circ}\cos 25^{\circ}} equal to?
- A. 1
- B. \sqrt{3}
- C. 2
- D. 4 ✓
Correct Answer: D. 4
Explanation
Multiply and divide the numerator by 2: 2(\frac{\sqrt{3}}{2}\cos 10^{\circ} - \frac{1}{2}\sin 10^{\circ}) = 2(\sin 60^{\circ}\cos 10^{\circ} - \cos 60^{\circ}\sin 10^{\circ}) = 2\sin(60^{\circ} - 10^{\circ}) = 2\sin 50^{\circ}. The denominator is \sin 25^{\circ}\cos 25^{\circ} = \frac{1}{2}(2\sin 25^{\circ}\cos 25^{\circ}) = \frac{1}{2}\sin 50^{\circ}. Thus, the fraction is \frac{2\sin 50^{\circ}}{\frac{1}{2}\sin 50^{\circ}} = 4.