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Q.

lf p=cos 2α+i sin 2α, q=cos 2β+i sin 2β then p/q-q/p =

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a

2i sinα-β

b

2 sinα-β

c

2i cosα-β

d

2 cosα-β

answer is A.

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Detailed Solution

p=cos2α+i sin n2α

p=cos α+i sin α2

 p=cos α+i sin α

q=cos2β+i sin2β=cos β+i sin β2

q=cos β+i sin β

pq=cos α+i sin αcos β+i sin β=cis αcis β=cis(α-β)

pq=cos(α-β)+i sin(α-β)

qp=cos(α-β)-i sin(α-β)

  pq-qp(cos(α-β)+i sin(α-β))-(cos(α-β)-i sin(α-β))

 =2 i sin (α-β)

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