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Q.

The value of (cos41°+cos42°+cos43°+....+cos4179°)(sin41°+sin42°+sin43°+....+sin4179°)=___ 

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a

2sin1°

b

–1

c

2cos1°

d

0

answer is B.

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Detailed Solution

 cos4θsin4θ=(cos2θ+sin2θ)(cos2θsin2θ)=1×cos2θ
 cos2+cos4+cos6+.....+cos358
 2(cos2+cos4+.....+cos178)+cos180
 n=89,α=2,β=2
 2sin89sin1°cos90°+(1)=1 

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