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Q.

If AB=0, then for the matrices A=cos2θcosθsinθcosθsinθsin2θ and B=cos2ϕcosϕsinϕcosϕsinϕsin2ϕ  θ-ϕ is

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a

an odd multiple of π2

b

an odd multiple of π

c

an even multiple of π2

d

0

answer is A.

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Detailed Solution

We have AB=cos2θcosθsinθcosθsinθsin2θcos2ϕcosϕsinϕcosϕsinϕsin2ϕ     =cos2θcos2ϕ+cosθsinθcosϕsinϕcos2θcosϕsinϕ+cosθsinθsin2ϕcosθsinθcos2ϕ+sin2θcosϕsinϕcosθsinθcosϕsinϕ+sin2θsin2ϕ       =cos(θ−ϕ)cosθcosϕcosθsinϕsinθcosϕsinθsinϕ Since AB=0,∴cos(θ−ϕ)=0∴(θ−ϕ) is an odd multiple of π2
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If AB=0, then for the matrices A=cos2θcosθsinθcosθsinθsin2θ and B=cos2ϕcosϕsinϕcosϕsinϕsin2ϕ  θ-ϕ is