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Question

Let S=2x1x2+2y1y2+2z1z2 and
T=8xyz(1x2)(1y2)(1z2), where x,y,z±1. If x+y+z=xyz, then the value of ST is

A
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B
0
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C
1
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D
2
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Solution

The correct option is B 0
Let x=tanA,y=tanB,z=tanC
S=2x1x2+2y1y2+2z1z2S=2tanA1tan2A+2tanB1tan2B+2tanC1tan2CS=tan2A+tan2B+tan2C(1)

T=8xyz(1x2)(1y2)(1z2)T=2tanA1tan2A×2tanB1tan2B×2tanC1tan2CT=tan2Atan2Btan2C

Given
x+y+z=xyztanA+tanB+tanC=tanAtanBtanC
So,
A+B+C=nπ, nZ2A+2B+2C=2nπ, nZtan(2A+2B+2C)=0tan2A+tan2B+tan2C =tan2Atan2Btan2CS=TST=0

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