A) \[x+y+z-xyz=0\]
B) \[x+y+z+xyz=0\]
C) \[xy+yz+zx+1=0\]
D) \[xy+yz+zx-1=0\]
Correct Answer: D
Solution :
Given that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\frac{\pi }{2}\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left[ \frac{x+y+z-xyz}{1-xy-yz-xz} \right]=\frac{\pi }{2}\] \[\Rightarrow \,\,\left[ \frac{x+y+z-xyz}{1-xy-yz-zx} \right]=\tan \frac{\pi }{2}=\frac{1}{0}\] Hence \[xy+yz+zx-1=0\]. Trick : \[x=y=z=\frac{1}{\sqrt{3}},\] so that \[{{\tan }^{-1}}\frac{1}{\sqrt{3}}+{{\tan }^{-1}}\frac{1}{\sqrt{3}}+{{\tan }^{-1}}\frac{1}{\sqrt{3}}=\frac{\pi }{2}\] Obviously (d) holds for these values of x, y, z.
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