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Question

If A+B+C=180, then sin2A2+sin2B2+sin2C2=

A
1+2sinA2sinB2sinC2
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B
1+2cosA2cosB2cosC2
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C
12sinA2sinB2sinC2
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D
12cosA2cosB2cosC2
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Solution

The correct option is D 12sinA2sinB2sinC2
Let sin2A2+sin2B2+sin2C2=I
I=sin2A2+(1cos2B2)+sin2C2
=sin2A2cos2B2+sin2C2+1
=cos(A+B2)cos(AB2)+sin2C2+1
As A+B+C=180A+B2=90C2
I=cos(90C2)cos(AB2)+sin2C2+1
=sin(C2)cos(AB2)+sin2C2+1
=sin(C2)(cos(AB2)sin(C2))+1
Now using C2=90A+B2
I=sin(C2)(cos(AB2)sin(90oA+B2))+1
=sin(C2)(cos(AB2)cos(A+B2))+1
=2sin(A2)sin(B2)sin(C2)+1
Hence, option C is correct.

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