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a
2cot2012(A+B2)
b
2cot2012(A−B2)
c
2tan2012(A−B2)
d
0
answer is B.
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Detailed Solution
cosA+cosBsinA-sinB=2cosA+B2.cosA-B22cosA+B2.sinA-B2=cotA-B2, sinA+sinBcosA-cosB=2sinA+B2.cosA-B2-2sinA+B2.sinA-B2=-cotA-B2since n is even, so required answer is2cot2012(A−B2)