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Question

If A+B+C=90, then sin2A+sin2B+sin2C=?

A
1+4sinAsinBsinC
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B
12sinAsinBsinC
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C
2+2sinAsinBsinC
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D
4sinAsinBcosC
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Solution

The correct option is C 12sinAsinBsinC
Given

A+B+C=90

sin2A+sin2B+sin2C

1cos2A+sin2B+sin2C

1(cos2Asin2B)+sin2C

1cos(A+B)cos(AB)+sin2C

((cos2Asin2B)=cos(A+B)cos(AB))
1cos(90C)cos(AB)+sin2C

1sin(C)cos(AB)+sin2C

1sinC(cos(AB)+sinC)

1sinC(cos(AB)+sin(90(A+B)))

1sinC(cos(AB)+cos(A+B))

(cosC+cosD=2sin(C+D2)sin(CD2))

12sinAsinBsinC

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