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Questions  

If A+B+C=π/2, then sin2A+sin2B+sin2C+2sinAsinBsinC is equal to 

a
0
b
1
c
-1
d
none of these

detailed solution

Correct option is B

sin2⁡A+sin2⁡B+sin2⁡C+2sin⁡Asin⁡Bsin⁡C=1−cos2⁡A−sin2⁡B+sin⁡C[sin⁡C+2sin⁡Asin⁡B]=1−cos⁡(A+B)cos⁡(A−B)+sin⁡C[cos⁡(A+B) +2sin⁡Asin⁡B]      [∵C=π/2−A−B]=1−cos⁡(π/2−C)cos⁡(A−B)+sin⁡Ccos⁡(A−B)=1

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