(sin75^0 + sin15^0)/(sin75^0 - sin15^0) =? 

Explanation:

Another Explanation (5): \( \frac{\sin 75^\circ + \sin 15^\circ}{\sin 75^\circ - \sin 15^\circ} = ? \) আমরা জানি, \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \) এবং \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \) সুতরাং, \( \sin 75^\circ + \sin 15^\circ = 2 \sin \frac{75^\circ + 15^\circ}{2} \cos \frac{75^\circ - 15^\circ}{2} = 2 \sin 45^\circ \cos 30^\circ \) এবং \( \sin 75^\circ - \sin 15^\circ = 2 \cos \frac{75^\circ + 15^\circ}{2} \sin \frac{75^\circ - 15^\circ}{2} = 2 \cos 45^\circ \sin 30^\circ \) অতএব, \( \frac{\sin 75^\circ + \sin 15^\circ}{\sin 75^\circ - \sin 15^\circ} = \frac{2 \sin 45^\circ \cos 30^\circ}{2 \cos 45^\circ \sin 30^\circ} = \frac{\sin 45^\circ}{\cos 45^\circ} \cdot \frac{\cos 30^\circ}{\sin 30^\circ} = \tan 45^\circ \cdot \cot 30^\circ \) আমরা জানি, \( \tan 45^\circ = 1 \) এবং \( \cot 30^\circ = \sqrt{3} \) সুতরাং, \( \frac{\sin 75^\circ + \sin 15^\circ}{\sin 75^\circ - \sin 15^\circ} = 1 \cdot \sqrt{3} = \sqrt{3} \) সুতরাং, উত্তর: \( \sqrt{3} \) 🎉