Prove: 2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=cot⁡2(x2)\frac{2 \sin x + \sin 2x}{2 \sin x - \sin 2x} = \cot^2 \left( \frac{x}{2} \right)

Mathematics

Problem

Prove: 2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=cot⁡2(x2)\frac{2 \sin x + \sin 2x}{2 \sin x - \sin 2x} = \cot^2 \left( \frac{x}{2} \right)

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