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Question

If xcosθ=ycos(θ+2π3)=zcos(θ+4π3), then xy+yz+zx=0.

A
True
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B
False
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Solution

The correct option is A True
xy+yz+zx=xyz(1x+1y+1z)
Put xcosθ=ycos(θ+2π3)=zcos(θ+4π3)=k(say)
Then
x=kcosθ,y=kcos(θ+2π3) and z=kcos(θ+4π3)
So that 1x+1y+1z
=1k[cosθ+cos(θ+2π3)+cos(θ+4π3)]
=1k[cosθ+cos(θ+ππ3)+cos(θ+π+π3)]
=1k[cosθcos(θπ3)cos(θ+π3)]
=1k[cosθ(cosθcosπ3+sinθsinπ3)(cosθcosπ3sinθsinπ3)]
=1k[cosθcosθcosπ3sinθsinπ3cosθcosπ3+sinθsinπ3]
=1k[cosθ2cosθcosπ3]
=1kcosθ(12cosπ3)
=1kcosθ(12×12)
=1kcosθ(11)
=1kcosθ×0=0
Hence the above statement is true.

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