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Question

# lf A+B+C=180∘. Match the followingList -I List -II1. cosA+cosB+cosC a.2+2cosAcosBcosC2. sin2A+sin2B+sin2C b.4cosA2cosB2cosC23. sin2A+sin2B+sin2C c.1+4sinA2sinB2sinC24.sinA+sinB+sinC d. 4sinAsinBsinC

A
1c,2a,3d,4b
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B
1d,2b,3a,4c
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C
1a,2c,3b,4d
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D
1b,2d,3c,4a
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Solution

## The correct option is A 1−c,2−a,3−d,4−bA) cosA+cosB+cosC=2cos(A+B2)cos(A−B2)+cos(π−(A+B))=2cos(A+B2)cos(A−B2)−cos(A+B)=2cos(A+B2)cos(A−B2)−2cos2(A+B2)+1=2cos(A+B2)(cos(A−B2)−cos(A+B2))+1=2cos(π2−c2)(2sinA2sinB2)+1=1+4sinA2sinB2sinC2B) sin2A+sin2B+sin2C=1−cos2A2+1−cos2B2+1−cos2B2=32−12(cos2A+cos2B+cos2C)=32−12(2cos(A+B)cos(A−B)+cos(2π−2(A+B)))=32−12(2cos(A−B)cos(A−B)+cos2(A+B)−1)=2+2cosAcosBcosCC) sin2A+sin2B+sin2C=2sin(A+B)cos(A−B)+sin(2π−2(A+B))=2sin(A+B)cos(A−B)−sin2(A+B)=2sin(A+B)cos(A−B)−2sin(A+B)cos(A−B)=2sin(A+B)(cos(A−B))−cos(A+B)=4sinAsinBsinCD) sinA+sinB+sinC=2sin(A+B2)cos(A−B2)+sin(A+B)=2sin(A+B2)cos(A−B2)+2sin(A+B2)cos(A+B2)=2sin(A+B2)(cos(A−B2)+cos(A+B2))=4cosA2cosB2cosC2

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