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Q.

If A+B+C+D=2π, then cos2A+cos2B+cos2C+cos2D=

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a

4cos(A+B)cos(A+C)cos(A+D)

b

4cos(AB)cos(A+C)cos(A+D)

c

4cos(A+B)cos(AC)cos(A+D)

d

4cos(A+B)cos(A+C)cos(AD)

answer is A.

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Detailed Solution

=2cos(A+B)cos(AB)+2cos(C+D)cos(CD)=2cos(A+B)[cos(AB)+cos(CD)]=2cos(A+B)2cosAB+CD2cosABC+D2=4cos(A+B)cosA+C[2π(A+C)]2cosA+D[2π(A+D)]2=4cos(A+B)cos(A+C)cos(A+D)

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