What is the value of \sin \theta+\cos \theta, if \theta satisfies the equation \cot^{2}\theta-(\sqrt{3}+1)\cot \theta+\sqrt{3}=0; 0 \lt \theta \lt \frac{\pi}{4}?
- A. \sqrt{2}
- B. 2
- C. \frac{\sqrt{3}+1}{2} ✓
- D. \frac{\sqrt{3}-1}{2}
Correct Answer: C. \frac{\sqrt{3}+1}{2}
Explanation
Factorizing the quadratic equation: (\cot\theta - \sqrt{3})(\cot\theta - 1) = 0. So \cot\theta = \sqrt{3} or \cot\theta = 1. Since 0 \lt \theta \lt \frac{\pi}{4}, \cot\theta \gt 1. Therefore, \cot\theta = \sqrt{3}, meaning \theta = 30^\circ. Then \sin 30^\circ + \cos 30^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}+1}{2}.
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