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Q.

If (1+cosθ−isinθ)(1+cos2θ+isin2θ)=x+iy then x2+y2=

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a

16cos2θsin2(θ2)

b

16sin2θcos2(θ2)

c

16sin2θsin2(θ2)

d

16cos2θcos2(θ2)

answer is D.

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Detailed Solution

((1+cosθ)−isinθ)((1+cos2θ)+isin2θ) =(2cos2θ2−i2sinθ2cosθ2)(2cos2θ+i2sinθcosθ)=2cosθ22cosθ[cosθ2−isinθ2][cosθ+isinθ]x+iy=4cosθ2cosθ(cosθ2+isinθ2)Taking modulus both side and squaring∴x2+y2=16cos2θcos2θ2(cos2θ2+sin2θ2)=16cos2θcos2θ2
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