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Q.

(cosA+cosBsinA−sinB)2012+(sinA+sinBcosA−cosB)2012=

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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)
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