If A+B+C+D=2π, then −4cos⁡A+B2sin⁡A+C2cos⁡A−D2=

Since A+B+C+D=2π, take A=B=C=D=π/2. Then −4cos⁡A+B2sin⁡A+C2cos⁡A−D2=0. ∴Answer 2 is correct

If A+B+C+D=2π, then −4cos⁡A+B2sin⁡A+C2cos⁡A−D2=

Since A+B+C+D=2π, take A=B=C=D=π/2. Then −4cos⁡A+B2sin⁡A+C2cos⁡A−D2=0. ∴Answer 2 is correct

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$If A + B + C + D = 2 π, then - 4 \cos (\frac{A + B}{2}) \sin (\frac{A + C}{2}) \cos (\frac{A - D}{2}) =$

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sin A + sin B + sin C + sin D

sin A - sin B + sin C + sin D

sin A - sin B + sin C - sin D

sin A + sin B + sin C - sin D

answer is B.

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$Since A + B + C + D = 2 π, take A = B = C = D = π / 2 .$
$Then - 4 \cos (\frac{A + B}{2}) \sin (\frac{A + C}{2}) \cos (\frac{A - D}{2}) = 0 .$ $∴$ Answer 2 is correct

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If A+B+C+D=2π, then −4cos⁡A+B2sin⁡A+C2cos⁡A−D2=

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$If A + B + C + D = 2 π, then - 4 \cos (\frac{A + B}{2}) \sin (\frac{A + C}{2}) \cos (\frac{A - D}{2}) =$