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B
−1−xyz
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C
1+xyz
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D
−1+xyz
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Solution
The correct option is C1+xyz If cot−1x+cot−1y+cot−1z=π4 cot−1x+cot−1y=π4−cot−1z Take cot on both sides. cot(cot−1x+cot−1y)=cot(π4−cot−1z).....[∵cot(A+B)=cotAcotB−1cotB+cotA] xy−1x+y=z+1(z−1) −xy+xyz−z+1=zx+yz+x+y ⇒xz+yz+xy+x+y+z=1+xyz