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Q.

In triangle ABC, sin⁡A+sin⁡B+sin⁡Csin⁡A+sin⁡B−sin⁡C=

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a

tan⁡A2cot⁡B2

b

cot⁡A2tan⁡B2

c

cot⁡A2cot⁡B2

d

tan⁡A2tan⁡B2

answer is C.

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Detailed Solution

Denominator = sin A + sin B - sin C                      =2sin⁡A+B2cos⁡A−B2−2sin⁡C2cos⁡C2=2cos⁡C2cos⁡A−B2−sin⁡C2=2cos⁡C2cos⁡A−B2−cos⁡A+B2=2cos⁡C22sin⁡A2sin⁡B2=4sin⁡A2sin⁡B2cos⁡C2Also  sin⁡A+sin⁡B+sin⁡C=4cos⁡A2cos⁡B2cos⁡C2⇒sin⁡A+sin⁡B+sin⁡Csin⁡A+sin⁡B−sin⁡C=cot⁡A2cot⁡B2
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In triangle ABC, sin⁡A+sin⁡B+sin⁡Csin⁡A+sin⁡B−sin⁡C=