sin⁡2A+sin⁡2B+sin⁡2Csin⁡A+sin⁡B+sin⁡C is equal to

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8sin⁡A2sin⁡B2sin⁡C2

8cos⁡A2cos⁡B2cos⁡C2

8tan⁡A2tan⁡B2tan⁡C2

8cot⁡A2cot⁡B2cot⁡C2

answer is A.

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Detailed Solution

sin⁡2A+sin⁡2B+sin⁡2Csin⁡A+sin⁡B+sin⁡C=4sin⁡Asin⁡Bsin⁡C4cos⁡A2cos⁡B2cos⁡C2=8sin⁡A2sin⁡B2sin⁡C2 ∵sin⁡A=2sin⁡A2cos⁡A2

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